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.