Latihan Soal 1 Persiapan PAS Gasal Matematika Wajib Kelas XI

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

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

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

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

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

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

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

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

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

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