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Q.
Let $n( > 1)$ be a positive integer. Then largest integer $m$ such that $\left(n^{m}+1\right)$ divides $1+n+n^{2}+\ldots+n^{255}$ is
Binomial Theorem
Solution:
We have, $S=1+n+n^{2}+\ldots+n^{255}$
$\Rightarrow S=\frac{1\left(n^{256}-1\right)}{n-1}=\left(n^{128}+1\right) \frac{n^{128}-1}{n-1}$
$\therefore S=\left(n^{128}+1\right)\left(1+n+n^{2}+\ldots+n^{127}\right)$
Thus, the largest value of $m$ for which $1+n+n^{2}$
$+\ldots+n^{255}$ is divisible by $n^{m}+1$ is $128$.