Question

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

• A. $32$
• B. $60$
• C. $65$
• D. $63$

Answer 1

The correct option is D

$63$

Explanation of the correct option.

Compute the required value:

Given: ${49}^{125}+{49}^{124}+......+{49}^2+49+1$

It can be written as, $1+49+{49}^2+...............+{49}^{124}+{49}^{125}$

It is a G.P. series with $a=1,r=49,n=126$

Since, the sum of G.P. is given by, $S=\frac{a(r^n-1)}{r-1}$

Substitute the values,

$$\begin{gathered} S=\frac{1({\left(49\right)}^{126}-1)}{49-1} \\ \Rightarrow S=\frac{{\left(49\right)}^{126}-1}{48} \\ \Rightarrow S=\frac{\left({49}^{63}-1\right)({49}^{63}+1)}{48} \end{gathered}$$

Since ${49}^K+1$ is a factor of the above sum,

Therefore $k=63$.

Hence, option $D$ is the correct answer.