The value of the greatest positive integer $k$, such that $49^k+1$ is a factor of $48\left(49^{125}+49^{124}+\ldots+49^2+49+1\right)$ is

$1+(1+3)+(1+3+5)+(1+3+5+7)+\ldots$ to 10 terms $=$

Assertion (A) : $1+\frac{2 \cdot 1}{3 \cdot 2}+\frac{2 \cdot 5}{3 \cdot 6} \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \frac{1}{8}+\ldots \infty=\sqrt[3]{4}$

Reason (R) : |x| < 1,(1-x) $=1+n x+\frac{n(n+1)}{1 \cdot 2} x^2$$+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^{3}+\ldots$

The correct answer is :