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The greatest positive integer k, for which 49^{k}+1 is a factor of the sum 49^{125}+49^{124}+....+49^{2}+49+1, is :
Option: 1 32
Option: 2 60
Option: 3 65
Option: 4 63

Answers (1)

best_answer

Sum of n-term of a GP

 

Let Sn be the sum of n terms of the G.P. with the first term ‘a’ and common ratio ‘r’. Then 

{S_n=a\left ( \frac{r^n-1}{r-1} \right )}

The above formula does not hold for r = 1, For r = 1 the sum of n terms of the G.P. is S_n = na.

 

Now,

\begin{align*} &1+49+49^2+......49^{125}\\ &\\\ &\Rightarrow \frac{1(49^{126}-1)}{48}\\ &\Rightarrow \frac{(49^{63}-1)(49^{63}+1)}{48}\\ \end{align*}

As xn - yn is always divisible by (x - y), so \frac{(49^{63}-1)}{48} is an integrer

Hence, K=63

Correct Option (4)

Posted by

Ritika Jonwal

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