Sequences And Series - Algebra - Mathematics Previous Year Questions

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 :

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 $=$

The value of the greatest integer $k$ satisfying the inequation $2^{n+4}+12 \geq k(n+4)$ for all $n \in N$ is

If $\frac{1}{2 \cdot 7}+\frac{1}{7 \cdot 12}+\frac{1}{12 \cdot 17}+\frac{1}{17 \cdot 22}+\ldots$ to 10 terms $=k$, then $k=$

$t_1, t_2, t_3, \ldots, t_n$ are positive integers, $S_n=t_1+t_2+t_3+\ldots+t_n$, $S_1=1^2, S_2=3^2, S_3=6^2, S_4=10^2, S_5=15^2$ and similarly other terms are there. Following this pattern, if $S_{10}=k^2$ then $k=$