Question

For all $n\in N,{10}^n+3\cdot 4^{n+2}+5$ is divisible by

• A. $54$
• B. $207$
• C. $9$
• D. $208$

Answer 1

Answer: (C)

$9$

Let $P\left(n\right)={10}^n+{3.4}^{n+2}+5.$ We have to show than $P\left(n\right)$ is divisible by $9$ for any value of $n\in N.$ We can do it by using principal of mathematical induction.

Step 1 : Let $n=1$

$\Rightarrow P\left(1\right)=10++{3.4}^3+5=207.$

$207$ is divisible by $9.$

$\therefore P\left(1\right)$ is true.

Step 2 : For $n=k$ we have to assume

$P\left(k\right)$ is true

$\Rightarrow P\left(k\right)={10}^k+{3.4}^{k+2}+5$ is divisible by $9.$

$\Rightarrow P\left(k\right)=9\times m$

$\Rightarrow {10}^k=9m-{3.4}^{k+2}+5$

Step 3 : we have to prove that $P(k+1)$ is true.

$\Rightarrow P(k+1)={10}^{k+1}+{3.4}^{k+3}+5$

$={10.10}^k+{3.4}^{k+3}.4+5$

$=10(9m-{3.4}^{k+3}-5)+{12.4}^{k+3}+5$

$=90m+4^{k+3}(12-30)-50+5$

$=90m-{18.4}^{k+3}-45$

$\Rightarrow P(k+1)=9(10m-{2.4}^{k+3}-5)$

$\Rightarrow P(k+1)=9\times n$

$\therefore P(k+1)$ is divisible by $9$ $\xi$ hence, true.

So by POMI $P\left(n\right)$ is true fr all $n\in N.$