CBSE BoardEnglish MediumSTD 11 ScienceMathsPrinciple of Mathematical Induction1 Mark
MCQ
For all $n \in N, 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by:
A
$19$
✓
$17$
C
$23$
D
$25$
✓
Answer
Correct option: B.
$17$
Let $\text{P(n): }3\times5^{2\text{n}+1}+2^{3\text{n}+1}$
For $\text{P}(1):\ 3\times5^{2\times1+1}+2^{3\times1+1}=3.5^3+2^4$
$=3\times(125)+16=375+16$
$=391=23\times17$
So $,\text{P}(1)$ is divisible by $17.$
Assume $\text{P}((k)$ is divisible by $17\Rightarrow\text{P(k): }3\times5^{2\text{k}+1}+2^{3\text{k}+1}=17\lambda_1,\lambda_1\in\text{N}$
So$, 3\times5^{2\text{k}+1}=17\lambda_1-2^{3\text{k}+1}\ ....(\text{i})$
Now we have to prove $P(k + 1)$ is divisible by $17\Rightarrow\text{P(k+1): }3\times5^{2\text{k}+3}+2^{3\text{k}+4}=17\lambda_2,\lambda_2\in\text{N}$
$=\big[17\lambda_1-2^{3\text{k}+1}\big]\times5^2+2^{3\text{k}+1}\cdot2^{3}$
$=17\lambda_1\cdot25-25\cdot2^{3\text{k}+1}+8\cdot2^{3\text{k}+1}$
$=17\lambda_1\cdot25-17\cdot2^{3\text{k}+1}$
$=17\big[25\lambda_1-2^{3\text{k}+1}\big]$
$=17\lambda_2,\lambda_2=25\lambda_1-2^{3\text{k}+1}$
So$, \text{P}((k + 1)$ is divisible by $17$ whenever $\text{P}((k)$ is divisible by $17.$
Hence by the principle of mathematical induction we have $\text{P}((n): 3 \times 5^{2n+1} + 2^{3n+1}$ is divisible by $17,\forall\text{n}\in\text{N}$
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