Browse Questions

# $49^n+16n-1$ is divisible by

$\begin{array}{1 1}(A)\;3\\(B)\;19\\(C)\;64\\(D)\;29\end{array}$

Toolbox:
• $(a+b)^n=nC_0a^nb^1+nC_1a^{n-1}b^2.........nC_n a^0b^n$
$(49)^n+16n-1$
$\Rightarrow (1+48)^n+16n-1$
$\Rightarrow 1+48n+.....48^n+16n-1$
$\Rightarrow 64n+nC_2(48)^2+nC_3(48)^3+......+(48)^n$
$\Rightarrow 64(n+nC_2(6)^2+nC_3(6)^348+......+(6)^n8^{n-2}$
$\therefore 49^n+16n-1$ is divisible by 64
Hence (C) is the correct answer.