Question

If $P\left(n\right):{49}^n+16n+k$ is divisible by $64$ for $n\in N$ is true, then the largest negative integral value of $k$ is
(use principle of mathematical induction)

• A. $-2$
• B. $-1$
• C. $-3$
• D. $-4$

Answer 1

The correct option is B $-1$

As, $P\left(n\right)$ is true for $n\in N$
Also true for $n=1,P\left(1\right):65+k$ is divisible by $64$
$64={49}^1+16\times 1+k=65+k$
Thus $k=-1,-65$
So, largest negative integral value of $k=-1$
Let $P\left(n\right):{49}^n+16n-1$ is divisible by $64$ is true for $k=n$
$\Rightarrow {49}^n+16n-1=64p,p\in Z$
check at $k=n+1,P(n+1):{49}^{n+1}+16(n+1)-1$
$\Rightarrow P(n+1)=49(64p-16n+1)+16n+15=64\times 49p-48\times 16n+64$ is divisible by $64$ is true.
So, $P\left(n\right):{49}^n+16n-1$ is divisible by $64.$