Question

Let n > 1, be a positive integer. Then the largest integer m, such that $(n^m+1)$ divides $(1+n+n^2+n^3+...+n^{127})$ is :

• A. 127
• B. 63
• C. 64
• D. 32

Answer 1

The correct option is C 64

$LetS=1+n+n^2+...+n^{127}$
$=\frac{1(n^{128}-1)}{(n-1)}$ $[\because S_n=\frac{a(r^n-1)}{r-1};r>1]$
$=\frac{n^{128}-1}{(n-1)}=\frac{(n^{64}+1)(n^{64}-1)}{(n-1)}$
Thus at m = 64 given expression is divisible by $(n^m+1)$