Contoh Soal 5 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 21.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 3+6+12+24+...+\left({3.2}^{n-1}\right)=3.(2^n-1)\\ & \text{Jika}S(n)\text{benar, untuk}n=k+1,\text{maka}\\ & \text{ruas kiri persamaan tersebut dapat dituliskan}\\ & \text{dengan}....\\ & \begin{array}{l}\\ \text{a}.& 3+6+12+24+...+{3.2}^{k+1}\\ \text{b}.& 3+6+12+24+...+{3.2}^{k-1}\\ \text{c}.& 3+6+12+24+...+{3.2}^{k-1}+{3.2}^k\\ \text{d}.& 3+6+12+24+...+{3.2}^{k-1}+{3.2}^{k+1}\\ \text{e}.& 3+6+12+24+...+{3.2}^k+{3.2}^{k+1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& 3+6+12+24+...+{3.2}^{n-1}=3.(2^n-1)\\ & 3+6+12+24+...+{3.2}^{k-1}+{3.2}^k=3.(2^{k+1}-1)\end{array}\end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S. 2018. Bupena Matematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: ERLANGGA.

Contoh Soal 4 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 16.& \text{Diketahui}1+5+9+...+(4n-1)=2n^2-n\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}m<k\text{dengan}\\ & m,k\text{bilangan asli juga},\text{maka}\\ & (4m-3)+(4m+1)+...+(4k-3)=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}& 1+5+9+...+(4m-3)+(4m+1)+...+(4k-3)\\ & =\underset{2k^2-k}{\underbrace{1+5+...+(4k-3)}}-\underset{2(m-1{)}^2-(m-1)}{\underbrace{1+5+...+(4(m-1)-3)}}\\ & =2k^2-k-(2(m-1{)}^2-(m-1))\\ & =2k^2-k-2(m-1{)}^2+(m-1)\\ & =2k^2-k-2(m^2-2m+1)+m-1\\ & =2k^2-k-2m^2+4m-2+m-1\\ & =2k^2-k-2m^2+5m-3\\ & =(k-m+1)(2k+2m-3)\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Diketahui}{2}^1+2^2+2^3+...+2^n=2^{n+1}-2\\ & \text{dengan}n\text{bilangan asli}.\text{Jika}k\text{bilangan asli},\\ & \text{maka}{2}^2+2^3+2^4+...+2^k+2^{k+1}=....\\ & \begin{array}{l}\\ \text{a}.& (k-m)(2k+2m-2)\\ \text{b}.& (k-m+1)(2k+2m-3)\\ \text{c}.& (k-m+1)(2k-2m+1)\\ \text{d}.& (k-m+1)(2k^2+2m^2-3)\\ \text{e}.& (k-m{)}^2(2k-2m+4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 2^2+2^3+2^4+...+2^k+2^{k+1}\\ & =2^1+2^2+2^3+2^4+...+2^k+2^{k+1}-2^1\\ & =\underset{2^{k+1+1}-2}{\underbrace{2^1+2^2+2^3+2^4+...+2^k+2^{k+1}}}-2^1\\ & =2^{k+2}-2-2\\ & =2^{k+2}-4\\ & =2^k{.2}^2-4\\ & =2^k\times 4-4\\ & =4(2^k-1)\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 2+5+10+17+...+(n^2+1)={\displaystyle \frac{1}{6}(n+1)(2n^2+n+6)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka}....\\ & \begin{array}{l}\\ \text{a}.& 2+5+10+17+...+(k^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{b}.& 2+5+10+17+...+(n^2+1)\\ & ={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{c}.& 2+5+10+17+...+(k^2+2)\\ & ={\displaystyle \frac{1}{6}(k+2)(2k^2+5k+9)}\\ \text{d}.& (k^2+1)={\displaystyle \frac{1}{6}(k+1)(2k^2+k+6)}\\ \text{e}.& (n^2+2)={\displaystyle \frac{1}{6}(n+1)(2n^2+5n+9)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& \text{Cukup jelas}\\ & \text{Tinggal mensubstitusikan dari}\\ & \text{tiap}n\text{diganti}k\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 12+17+22+...+(5n+7)={\displaystyle \frac{1}{2}(n+1)(5n+14)}\\ & \text{Jika}S(n)\text{benar, untuk}n=k,\text{maka benar}\\ & \text{untuk}n=k+1.\text{Pernyataan ini dapat}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ \text{b}.& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{c}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+1)(5k+19)}\\ \text{d}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+14)}\\ \text{e}.& 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 12+17+22+...+(5k+7)={\displaystyle \frac{1}{2}(k+1)(5k+14)}\\ & 12+17+22+...+(5(k+1)+7)\\ & ={\displaystyle \frac{1}{2}((k+1)+1)(5(k+1)+14)}\\ & 12+17+22+...+(5k+12)={\displaystyle \frac{1}{2}(k+2)(5k+19)}\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Diketahui bahwa}S(n)\text{adalah formula dari}\\ & 4+5+6+7+...+(n+3)<5n^2\\ & \text{Jika}S(n)\text{benar, untuk}n=k+1,\text{maka}\\ & \text{pernyataan ini dapat ditulis dengan}....\\ & \begin{array}{l}\\ \text{a}.& 4+5+6+...+(k+4)<5k^2\\ \text{b}.& 4+5+6+...+(k+3)<5k^2\\ \text{c}.& 4+5+6+...+(k+3)<5(k+1{)}^2\\ \text{d}.& 4+5+6+...+(k+4)<5(k^2+2k+1)\\ \text{e}.& 4+5+6+...+(k+4)<5(k+1)(k-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& 4+5+6+...+(n+3)<5n^2\\ & \text{Saat}n=k+1,\text{maka}\\ & 4+5+6+...+((k+1)+3)<5(k+1{)}^2\\ & =4+5+6+...+(k+4)<5(k^2+2k+1)\end{array}\end{array}$

Contoh Soal 3 Induksi Matematika (Matematika Wajib Kelas XI)

 $\begin{array}{l}\\ 11.& \text{Perhatikanlah pernyataan-pernyataan berikut}\\ & (1){\displaystyle \sum _{i=1}^5(5i+2)=4{\displaystyle \sum _{i=1}^{5}i+10}}\\ & (2){\displaystyle \sum _{i=1}^5(5i^2-i)=5{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}i}}\\ & (3){\displaystyle \sum _{i=1}^5(3i-4)=3{\displaystyle \sum _{i=1}^5{i}^2-4}}\\ & (4){\displaystyle \sum _{i=1}^5(i+7i^2)={\displaystyle \sum _{i=1}^{5}i-7\sum _{i=1}^{5}i}}\\ & \text{Pernyataan yang tepat ditunjukkan oleh}....\\ & \begin{array}{l}\\ \text{a}.& (1)\text{dan}(2)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (1)\text{dan}(4)\\ \text{d}.& (2)\text{dan}(3)\\ \text{e}.& (2)\text{dan}(4)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}(1)& {\displaystyle \sum _{i=1}^5(5i+2)=4{\displaystyle \sum _{i=1}^{5}i+\sum _{i=1}^{5}2}}\\ & =4{\displaystyle \sum _{i=1}^{5}i+5\times 2}\\ & =4{\displaystyle \sum _{i=1}^{5}i+10}\\ (2)& {\displaystyle \sum _{i=1}^5(5i^2-i)=5{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}i}}\\ (3)& {\displaystyle \sum _{i=1}^5(3i-4)=3{\displaystyle \sum _{i=1}^5{i}^2-\sum _{i=1}^{5}4}}\\ & =3{\displaystyle \sum _{i=1}^5{i}^2-5\times 4}\\ & =3{\displaystyle \sum _{i=1}^5{i}^2-20}\\ (4)& {\displaystyle \sum _{i=1}^5(i+7i^2)={\displaystyle \sum _{i=1}^{5}i+7\sum _{i=1}^{5}i}}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& \text{Hasil dari}{\displaystyle \sum _{i=1}^4{i}^2+\sum _{i=5}^6{i}^2}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 86}\\ \text{b}.& {\displaystyle 91}\\ \text{c}.& {\displaystyle 95}\\ \text{d}.& {\displaystyle 101}\\ \text{e}.& {\displaystyle 105}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=1}^4{i}^2+\sum _{i=5}^6{i}^2}& ={\displaystyle \sum _{i=1}^6{i}^2}\\ & =1^2+2^2+3^2+4^2+5^2+6^2\\ & =1+4+9+16+25+36\\ & =91\end{array}\end{array}$

$\begin{array}{l}\\ 13.& \text{Hasil dari}{\displaystyle \sum _{i=2}^5(4i^2-2i)\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 144}\\ \text{b}.& {\displaystyle 148}\\ \text{c}.& {\displaystyle 154}\\ \text{d}.& {\displaystyle 164}\\ \text{e}.& {\displaystyle 188}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \sum _{i=2}^5(4i^2-2i)}\\ & =({4.2}^2-2.2)+({4.3}^2-2.3)+({4.4}^2-2.4)+({4.5}^2-2.5)\\ & =12+30+56+90\\ & =188\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Bentuk}{11}^n-1\text{dengan}n\text{bilangan asli}\\ & \text{akan habis dibagi oleh}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 7}\\ \text{b}.& {\displaystyle 9}\\ \text{c}.& {\displaystyle 10}\\ \text{d}.& {\displaystyle 11}\\ \text{e}.& {\displaystyle 13}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Bentuk}& {11}^n-1\\ \text{untuk}& n=1\\ & ={11}^1-1\\ & =10\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Rumus yang tepat untuk pola}12,13,14,15,...\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle U_n=n+9}\\ \text{b}.& {\displaystyle U_n=n+10}\\ \text{c}.& {\displaystyle U_n=n+11}\\ \text{d}.& {\displaystyle U_n=2n+10}\\ \text{e}.& {\displaystyle U_n=2n+11}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{Bentuk}& 12,13,14,15,...\\ \text{untuk}& U_n=pn+q\\ 12& =p+q\\ 13& =2p+q\\ \text{akan}& \text{didapatkan}\\ & \{\begin{array}{ll}p& =1\\ q& =11\end{array}\\ \text{Sehing}& \text{ga}\\ U_n& =n+11\end{array}\end{array}$

Contoh Soal 2 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 6.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan juga bahwa}{7}^n-2^n\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =7^n-2^n\\ P(1)& =7^1-2^1\\ & =7-2=5\\ & \text{adalah bilangan yang habis dibagi 5}\end{array}\end{array}$

$\begin{array}{l}\\ 7.& \text{Diketahui bahwa}P(n)\text{rumus dari}\\ & 3+6+9+\cdots +3n={\displaystyle \frac{3}{2}n(n+1)}\\ & \text{maka langkah pertama dengan induksi matematika}\\ & \text{dalam pembuktian rumus tersebut adalah}....\\ & \begin{array}{l}\\ \text{a}.& P(n)\text{benar untuk}n=-1\\ \text{b}.& P(n)\text{benar untuk}n=1\\ \text{c}.& P(n)\text{benar untuk}n\text{bilangan bulat}\\ \text{d}.& P(n)\text{benar untuk}n\text{bilangan rasional}\\ \text{d}.& P(n)\text{benar untuk}n\text{bilangan real}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Langkah}& \text{awal yang harus ditunjukkan adalah}\\ n=1& \text{atau}P(1)\text{harus benar, yaitu}:\\ P(1)& =3.1(\mathit{\text{ruas kiri}})={\displaystyle \frac{3}{2}.1.(1+1)(\mathit{\text{ruas kanan}})=3}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Bila kita hendak membuktikan}{\displaystyle \sum _{i=1}^n={\displaystyle \frac{1}{2}n(n+1)}}\\ & \text{dengan induksi matematika}\\ & \text{maka untuk langkah}n=k+1\\ & \text{bentuk yang harus ditunjukkan adalah}...\\ & \begin{array}{l}\\ \text{a}.& 1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ \text{b}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ \text{c}.& 1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+2)}\\ \text{d}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+1)(k+2)}\\ \text{e}.& 1+2+3+\cdots +k+(k+1)={\displaystyle \frac{1}{2}(k+2)(k+3)}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}P(n)& =1+2+3+\cdots +n={\displaystyle \frac{1}{2}n(n+1)}\\ P(k)& =1+2+3+\cdots +k={\displaystyle \frac{1}{2}k(k+1)}\\ P(k+1)& =1+2+3+\cdots +k+(k+1)\\ & =\underset{{\displaystyle \frac{1}{2}k(k+1)}}{\underbrace{1+2+3+\cdots +k}}+(k+1)\\ & ={\displaystyle \frac{1}{2}k(k+1)+(k+1)=(k+1)\left({\displaystyle \frac{1}{2}k+1}\right)}\\ & =(k+1){\displaystyle \frac{1}{2}(k+2)}\\ & ={\displaystyle \frac{1}{2}(k+1)(k+2)}\end{array}\end{array}$

$\begin{array}{l}\\ 9.& \text{Jika}P(n)={\displaystyle \frac{n-1}{n+3},\text{maka}P(k+1)}\\ & \text{dinyatakan dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k-1}{k+3}}\\ \text{b}.& {\displaystyle \frac{k-1}{k+4}}\\ \text{c}.& {\displaystyle \frac{k}{k+4}}\\ \text{d}.& {\displaystyle \frac{k+1}{k+4}}\\ \text{e}.& {\displaystyle \frac{k+1}{k+5}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)=& {\displaystyle \frac{n-1}{n+3}}\\ P(k+1)& ={\displaystyle \frac{k+1-1}{k+1+3}}\\ & ={\displaystyle \frac{k}{k+4}}\end{array}\end{array}$

$\begin{array}{l}\\ 10.& \text{Jika}P(n)={\displaystyle \frac{n^2+1}{4},\text{maka}}\\ & \text{pernyataan untuk}P(k+1)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{k^2+2k+1}{4}}\\ \text{b}.& {\displaystyle \frac{k^2+2k+2}{4}}\\ \text{c}.& {\displaystyle \frac{k^2+2k+2}{5}}\\ \text{d}.& {\displaystyle \frac{k^2+2k+3}{5}}\\ \text{e}.& {\displaystyle \frac{k^2+2k+3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}P(n)& ={\displaystyle \frac{n^2+1}{4}}\\ P(k+1)& ={\displaystyle \frac{(k+1{)}^2+1}{4}}\\ & ={\displaystyle \frac{k^2+2k+2}{4}}\end{array}\end{array}$

Contoh Soal 1 Induksi Matematika (Matematika Wajib Kelas XI)

$\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \sum _{i=1}^{6}16i\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 306\\ \text{b}.& 314\\ \text{c}.& 326\\ \text{d}.& 336\\ \text{e}.& 402\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=1}^{6}16i}& =16.1+16.2+16.3+16.4+16.5+16.6\\ & =16+32+48+64+80+96\\ & =336\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Hasil dari}{\displaystyle \sum _{i=2}^9{i}^2\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 274\\ \text{b}.& 278\\ \text{c}.& 280\\ \text{d}.& 284\\ \text{e}.& 286\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \sum _{i=2}^9{i}^2}& =2^2+3^2+4^2+5^2+..+9^2\\ & =4+9+16+25+...+81\\ & =284\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Poa bilangan}12,14,16,18,20,...,(2n+10).\\ & \text{Nilai suku ke-100 adalah}....\\ & \begin{array}{l}\\ \text{a}.& 180\\ \text{b}.& 194\\ \text{c}.& 198\\ \text{d}.& 208\\ \text{e}.& 210\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{U}_n& =2n+10\\ U_{100}& =2\times 100+10\\ & =210\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa jika}\\ & 31+39+47+\cdots +8n+23=4n^2+27n\\ & \text{dengan}k,n\in \mathbb{N}\text{maka}\\ & 31+39+47+\cdots +8n+23+8k+31=....\\ & \begin{array}{l}\\ \text{a}.& 4k^2+27k\\ \text{b}.& 4k^2+35k\\ \text{c}.& 4k^2+35k+31\\ \text{d}.& 4k^2+35k+1\\ \text{e}.& 4k^2+35k+54\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}& \underset{4k^2+27k}{\underbrace{31+39+47+\cdots +8k+23}}+8k+31\\ & =4k^2+27k+8k+31\\ & =4k^2+35k+31\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Dengan Induksi Matematika untuk}n\in \mathbb{N}\\ & \text{dapat dibuktikan bahwa}n(n+1)(n+2)\\ & \text{akan habis dibagi oleh}\\ & \begin{array}{l}\\ \text{a}.& 4\\ \text{b}.& 5\\ \text{c}.& 6\\ \text{d}.& 7\\ \text{e}.& 8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}P(n)& =n(n+1)(n+2)\\ P(1)& =1(1+1)(1+2)=1.2.3\\ & \text{adalah bilangan yang habis dibagi 6}\end{array}\end{array}$

Notasi Sigma Lanjutan Induksi Matematika (Matematika Wajib Kelas XI)

$\text{A. Pendahuluan}$

Notasi sigma dari asalnya dari yaitu dari huruf yunani yang memiliki makna jumlah. Dalam matematika lambang notasi sigma $"\sum "$  selanjutnya akan menunjukkan penjumlahan yang teratur sehingga penulisan sebuah deret dari suatu bilangan yang berpola tertentu dapat disederhanakan lebih ringkas.

Sebagai ilustrasinya untuk deretny adalah sebagai berikut

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100\\ \text{b}.1+3+5+7+9+\cdots +199\\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2\end{array}$

Dari bentuk deret di atas jika dimodelkan dengan notasi sigma maka bentuknya akan menjadi lebih sederhana, yaitu:

$\begin{array}{rl}& \sum _{i=1}^n{a}_i=a_1+a_2+a_3+\cdots +a_n\\ & \text{Dibaca}:"\text{Jumlah}\text{dari}{a}_i\text{untuk}i\\ & \text{dari 1 sampai dengan}n"\text{dan}{a}_i\text{adalah suku ke}-i\end{array}$

Sehingga contoh ilustrasi deret di atas jika dinotasikan dengan notasi sigma menjadi

$\begin{array}{l}\\ \text{a}.1+2+3+4+5+\cdots +100=\sum _{i=1}^{100}i\\ \\ \text{b}.1+3+5+7+9+\cdots +199=\sum _{i=1}^{100}(2i-1)\\ \\ \text{c}.1^2+2^2+3^2+4^2+5^2+\cdots +{100}^2=\sum _{i=1}^{100}{i}^2\end{array}$

$\text{B. Sifat-Sifat Notasi Sigma}$

Misalkan diketahui $a_k$  dan $b_k$  adalah suku ke-k dan C adalah sebuah konstanta, maka

$\begin{array}{l}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C=nC}\\ & ⧫{\displaystyle \sum _{k=1}^{n}C.a_k=C\sum _{k=1}^n{a}_k}\\ & ⧫{\displaystyle \sum _{k=1}^n(a_k+b_k)={\displaystyle \sum _{k=1}^n{a}_k+\sum _{k=1}^n{b}_k}}\\ & ⧫{\displaystyle \sum _{k=1}^n{(a_k+b_k)}^2={\displaystyle \sum _{k=1}^n{a}_k^2+2\sum _{k=1}^n{a}_k{b}_k+\sum _{k=1}^n{b}_k^2}}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^{n-1}{a}_k+a_n}\\ & ⧫{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=1}^m{a}_k+\sum _{k=m+1}^n{a}_k,1<m<n}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Uraikan jumlah berikut dengan lengkap}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k}& \text{f}.{\displaystyle \sum _{k=1}^3{2}^k}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)}& \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k}& \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)}& \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)}& \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)}\end{array}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & \begin{array}{l}\\ 1.& \text{Perhatikanlah,}\\ & \begin{array}{l}\\ & \text{a}.{\displaystyle \sum _{k=1}^{4}k=1+2+3+4=10}\\ & \text{b}.{\displaystyle \sum _{k=1}^4(k-3)=(1-3)+(2-3)+(3-3)+(4-3)=-2.}\\ & \text{c}.{\displaystyle \sum _{k=1}^{4}5k=5.1+5.2+5.3+5.4=50}\\ & \text{d}.{\displaystyle \sum _{k=1}^3(4k+2)=(4.1+2)+(4.2+2)+(4.3+2)=30}\\ & \text{e}.{\displaystyle \sum _{k=1}^3(2k+3)=(2.1+3)+(2.2+3)+(2.3+3)=21}\\ & \text{f}.{\displaystyle \sum _{k=1}^3{2}^k=2^1+2^2+2^3+2^4=2+4+8+16=30}\\ & \text{g}.{\displaystyle \sum _{k=1}^3\frac{1}{3^k}={\displaystyle \frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}={\displaystyle \frac{9+3+1}{27}=\frac{13}{27}}}}\\ & \text{h}.{\displaystyle \sum _{k=1}^5(k^2+1)=\cdots +\cdots +\cdots +\cdots +\cdots }\\ & \text{i}.{\displaystyle \sum _{k=1}^{4}5{\left(\frac{2}{3}\right)}^k=\cdots +\cdots +\cdots +\cdots }\\ & \text{j}.{\displaystyle \sum _{k=1}^5(k^2+2k-3)=\cdots +\cdots +\cdots +\cdots +\cdots }\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Nyatakanlah penjumlahan berikut dengan notasi sigma}\\ & \text{a}.2+4+8+16+32+64\\ & \text{b}.2+6+18+54+162\\ & \text{c}.15+24+35+48\\ & \text{d}.{\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}}\\ & \text{e}.ab+a^2{b}^2+a^3{b}^3+a^4{b}^4\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a})2+4+8+16+32+64={\displaystyle \sum _{k=1}^6{2}^k}\\ & (\text{b})2+6+18+54+162={\displaystyle \sum _{k=1}^5{2.3}^{k-1}}\\ & (\text{c})15+24+35+48={\displaystyle \sum _{k=1}^4(k^2+6k+8)}\\ & (\text{d}){\displaystyle \frac{2}{3}+\frac{4}{5}+\frac{8}{7}+\frac{16}{9}+\frac{32}{11}={\displaystyle \sum _{k=1}^5\frac{2^k}{(2k+1)}}}\\ & (\text{e})ab+a^2{b}^2+a^3{b}^3+a^4{b}^4={\displaystyle \sum _{k=1}^4(ab{)}^k}\end{array}$

$\begin{array}{l}\\ 3.& \text{Dengan menggunakan kaidah notasi sigma},\\ & \text{tunjukkan bahwa}\\ & \text{a}.{\displaystyle \sum _{k=1}^6(2k+3)=2\sum _{k=1}^{6}k+18}\\ & \text{b}.{\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=1}^{6}k+30}\\ & \text{c}.{\displaystyle \sum _{k=2}^5(2k^2+3k+3)=2\sum _{k=1}^4{k}^2+7\sum _{k=1}^{4}k+32}\\ & \text{d}.{\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=1}^6{k}^2-2\sum _{k=1}^{6}k+6}\\ & \text{e}.{\displaystyle \sum _{k=3}^6(k^2+2k-3)=\sum _{k=1}^6{k}^2+6\sum _{k=1}^4+20}\end{array}$

$\begin{array}{rl}& \text{Jawab}\\ & (\text{a}){\displaystyle \sum _{k=1}^6(2k+3)=\sum _{k=1}^{6}2k+\sum _{k=1}^{6}3}\\ & =\sum _{k=1}^{6}2k+6.3=2\sum _{k=1}^{6}k+18\\ & (\text{b}){\displaystyle \sum _{k=3}^8(k+3)=\sum _{k=3-2}^{8-2}((k+2)+3)}\\ & =\sum _{k=1}^6(k+5)=\sum _{k=1}^{6}k+\sum _{k=1}^{6}5=\sum _{k=1}^{6}k+6.5\\ & =\sum _{k=1}^{6}k+30\\ & (\text{c}){\displaystyle \sum _{k=2}^5(2k^2+3k+3)}\\ & =\sum _{k=2-1}^{5-1}(2(k+1{)}^2+3(k+1)+3)=\cdots \\ & (\text{d}){\displaystyle \sum _{k=0}^5{k}^2=\sum _{k=0+1}^{5+1}(k-1{)}^2=\cdots }\\ & (\text{e}){\displaystyle \sum _{k=3}^6(k^2+2k-3)=\cdots }\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{l}\\ .& \text{Buktikanlah bahwa}\\ & \text{a}.{\displaystyle \sum _{k=6}^{12}{k}^2=\sum _{k=1}^7{k}^2+10\sum _{k=1}^{7}k+175}\\ & \text{b}.{\displaystyle \sum _{k=1}^n(3k-1{)}^2=9\sum _{k=1}^n{k}^2-6\sum _{k=1}^{n}k+n}\\ & \text{c}.{\displaystyle \sum _{k=m}^n{a}_k=\sum _{k=m+p}^{n+p}{a}_{k-p}}\\ & \text{d}.{\displaystyle \sum _{i=m}^n{a}_i=\sum _{i=1}^n{a}_i-\sum _{i=1}^{m-1}{a}_i}\\ & \text{e}.{\displaystyle \sum _{k=1}^n{a}_k=\sum _{k=0}^{n-1}{a}_{k+1}=\sum _{k=2}^{n+2}{a}_{k-1}}\\ & \text{f}.{\displaystyle \sum _{k=1}^{n-5}{a}_k=\sum _{k=1}^n{a}_k-\sum _{k=(n-5)+1}^n{a}_k}\end{array}$

DAFTAR PUSTAKA

1. Kuntarti, Sulistiyono, Kurnianingsih, S. 2005. Matematika untuk SMA dan MA Kelas XII Program Ilmu Alam. Jakarta: Gelora Aksara Pratama.
2. Kuntarti, Sulistiyono, Kurnianingsih, S. 2007. Matematika SMA dan MA untuk Kelas XII Semester 2 Program IPA Standar Isi 2006. Jakarta: ESIS.

Induksi Matematika (Kelas XI Matematika Wajib)

$\text{A. Pendahuluan}$

Misalkan kita menjumlahkan 100 bilangan ganjil pertama (anggap saja sebagai penjumlahan suku pertama sampai suku ke seratus) yaitu : 1+3+5+...+199, maka untuk memudahkannya kita dapat menentukan cara menjumlahkan dengan atau menurut pola tertentu sebagaimana ilstrasi berikut ini

$\begin{array}{rl}1& =1^2=S_1\\ 1+3& =2^2=S_2\\ 1+3+5& =3^2=S_3\\ 1+3+5+7& =4^2=S_4\\ 1+3+5+7+9& =5^2=S_5\\ ⋮& \\ 1+3+5+7+9+\cdots +(2n-1)& =n^2=S_n\\ ⋮& \\ 1+3+5+7+9+\cdots +199& ={100}^2=S_{100}\end{array}$

$\text{B. Induksi Matematika}$

Pola bilangan tertentu dalam matematika sebagaimana misal contoh di atas dapat ditarik suatu bentuk umum. Selanjutnya untuk membuktikan bahwa suatu bentuk umum dari sebuah pernyataan berlaku, kita dapat menggunakan Induksi Matematika ini. Tentunya semunya dari pernyataan tersebut harus memenuhi kriteria tertentu. Sehingga Induksi Matematika dapat juga disebutkan sebagai proses pembuktian pernyataan (teorema) dari kejadian-kejadian khusus yang berlaku untuk setiap bilangan asli.

Dalam pembuktian dengan Induksi Matematika, perhatikanlah beberapa langkah-langkah ini

Misalkan $P(n)$ adalah suatu pernyataan yang akan dibuktikan kebenarannya untuk semua bilangan asli $n$, maka

$\begin{array}{rl}& \mathbf{\text{Langkahnya:}}\\ & (1)\text{buktikan}P(1)\text{benar untuk}n=1\\ & \text{Selanjutnya disebut Langkah Basis}\\ & (2)\text{Asumsikan pernyataan berlaku untuk}n=k,\\ & \text{yaitu}P(k)\text{benar, dengan}k\in A,\\ & \text{maka untuk}n=k+1\text{bahwa}P(k+1)\\ & \text{juga benar}\\ & \text{Selanjutnya disebut Langkah Induksi}\\ & (3)\text{Setelah langkah}(1)\text{dan}(2)\text{terpenuhi}\\ & \text{atau benar, maka dapat disimpulkan bahwa}\\ & P(n)\text{benar untuk setiap}n\\ & \text{Selanjutnya disebut Konkulsi}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{rl}(1)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}1={\displaystyle \frac{1(1+1)}{2}}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+2+3+...+k={\displaystyle \frac{k(k+1)}{2},\text{maka}}\\ & \text{untuk}n=k+1:\\ & 1+2+3+\cdots +(k)+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & \underset{{\displaystyle \frac{k(k+1)}{2}}}{\underbrace{1+2+3+\cdots +(k)}}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & {\displaystyle \frac{k(k+1)}{2}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)(\frac{k}{2}+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)\frac{k+2}{2}={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle \frac{(k+1)(k+2)}{2}={\displaystyle \frac{(k+1)(k+2)}{2}\equiv P(k+1)}}\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(2)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+3+5+7+...+(2n-1)=n^2\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+3+5+...+(2n-1)=n^2\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-1=1^2\iff 1=1\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+3+5+...+(2k-1)=k^2,\text{maka}\\ & \text{untuk}n=k+1:\\ & 1+3+5+\cdots +(2k-1)+(2(k+1)-1)=(k+1{)}^2\\ & \underset{{\displaystyle k^2}}{\underbrace{1+3+5+\cdots +(2k-1)}}+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2k+2-1)=(k+1{)}^2\\ & k^2+2k+1=(k+1{)}^2\\ & (k+1{)}^2=P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(3)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 2n-3<2^{n-2}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 2n-3<2^{n-2}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-3<2^{1-2}\\ & \text{demikian pula untuk}n=2\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (2k-3)<2^{k-2},\text{maka}\\ & \text{untuk}n=k+1:\\ & 2(k+1)-3=2k+2-3=(2k-3)+2<2^{k-2}+2\\ & \text{sehingga}\\ & (2k-3)+2<2^{k-2}+2<2^{k-2}+2^{k-2},\text{untuk}k\ge 3\\ & (2k-3)+2<{2.2}^{k-2}\\ & (2k-3)+2<2^{(k+1)-2}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(4)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & (1+h{)}^n\ge 1+nh\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv (1+h{)}^n\ge 1+nh\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}(1+h{)}^1\ge 1+1h\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (1+h{)}^k\ge 1+kh,\text{maka}\\ & \text{untuk}n=k+1:\\ & (1+h{)}^{k+1}\ge (1+kh)(1+h)\\ & (1+h{)}^{k+1}\ge (1+(k+1)h+k{h}^2)\\ & (1+h{)}^{k+1}\ge 1+(k+1)h\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}.& \mathbf{\text{Catatan}}\\ & \text{untuk}k=2\\ & (1+h{)}^2=1+2h+h^2\ge 1+2h,\text{maka}\\ & (1+h{)}^n\ge 1+nh\end{array}$

$\begin{array}{rl}(5)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena untuk}n=1\\ & 3^1={\displaystyle \frac{3}{2}(3^1-1)}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv \sum _{h=1}^k{3}^h={\displaystyle \frac{3}{2}(3^k-1),\text{maka}}\\ & \text{untuk}n=k+1:\\ & \sum _{h=1}^{k+1}{3}^h=\sum _{h=1}^k{3}^h+\sum _{h=k+1}^{k+1}{3}^h\\ & ={\displaystyle \frac{3}{2}(3^k-1)+3^{k+1}}\\ & ={\displaystyle \frac{1}{2}(3^{k+1}-3+{2.3}^{k+1})}\\ & ={\displaystyle \frac{1}{2}({3.3}^{k+1}-3)}\\ & ={\displaystyle \frac{3}{2}({.3}^{k+1}-1)}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{rl}.& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ \end{array}$

$.\begin{array}{l}\\ 1.& 2+4+6+8+...+2n=n^2+n\\ 2.& 1^2+2^2+3^2+...+n^2={\displaystyle \frac{1}{6}n(n+1)(2n+1)}\\ 3.& 1^3+2^3+3^3+...+n^3={\displaystyle \frac{1}{4}{n}^2(n+1{)}^2}\\ 4.& 1.2+2.3+3.4+...+n(n+1)={\displaystyle \frac{1}{3}n(n+1)(n+2)}\\ 5.& {\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}={\displaystyle \frac{n}{n+1}}}\\ 6.& {\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2n-1)(2n+1)}={\displaystyle \frac{n}{2n+1}}}\\ 7.& {\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}={\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}}}\\ 8.& 1+{\displaystyle \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n+1}-2}\\ 9.& n^3-n\text{habis dibagi oleh}3\\ 10.& n^5-n\text{habis dibagi oleh}5\\ 11.& 5^n+{6.7}^n+1\text{habis dibagi oleh}4\\ 12.& 5^{2n}-1\text{habis dibagi oleh}3\\ 13.& 3^n-1\ge 2^n\\ 14.& 2n+7<(n+3{)}^2\\ 15.& 2+4+6+8+...+2n\le 2^n\\ 16.& {(3+\sqrt{5})}^n+{(3-\sqrt{5})}^n\text{habis dibagi oleh}{2}^n\end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S., 2018. Bupena Mathematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: Erlangga.
2. Kemendikbud. 2017. Matematika untuk SMA/MA/SMK Kelas XI Edisi Revisi. Jakarta: Kementerian Pendidikan Nasional.
3. Tampomas, H. 1999. SeribuPena Matematika Jilid 3 untuk SMU Kelas 3. Jakarta: Erlangga
4. Tim ITB. 2007. Program Pembinaan Kompetensi Siswa Bidang Matematika Tahap 1. Bandung: LPPM ITB