Question

For all $n\in N;{49}^n+16n-1$ is divisible by

• A. $64$
• B. $49$
• C. $132$
• D. $32$

Answer 1

Answer: (A), (D)

$64$
$32$

$a={49}^n+16n-1=-1+16n+{(1+48)}^n$
$\Rightarrow a=-1+16n+(1+n_{C_1}48+n_{C_2}{48}^2+...+n_{C_n}{48}^n)=64n+n_{C_2}{48}^2+...+n_{C_n}{48}^n$
$\Rightarrow a=64(n+n_{C_2}{6}^2+n_{C_3}8{.6}^3+...+n_{C_n}{8}^{n-2}{.6}^n)$
Therefore, $a$ is divisible by $64,32$
Ans: A,D