Question:
The greatest positive integer $k$, for which $49^{k}+1$ is a factor of the sum $49^{125}+49^{124}+\ldots+49^{2}+49+1$, is:
Correct Option: , 2
Solution:
$\frac{(49)^{126}-1}{48}=\frac{\left((49)^{63}+1\right)\left(49^{63}-1\right)}{48}\left[\because S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}\right]$
$\therefore \mathrm{K}=63$
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