For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

If the coefficient of $x^{13}$ in the expansion of $\frac{(1+x)^2}{(1-2x)^3}$ is $A \times 2^{10}$,then $A=$

The sum of the series $1 + \frac{1 \times 3}{6} + \frac{1 \times 3 \times 5}{6 \times 8} + \dots \infty$ is

The binomial expansion $(7+3x)^{-2/5}$ is valid for all $x$ in the interval $\left(\frac{-7}{3}, \frac{7}{3}\right)$. If the $4^{th}$ term of its expansion is $kx^3$,then the value of $(7^{12/5}k)$ is:

$1+\frac{1}{4}+\frac{1 \cdot 3}{4 \cdot 8}+\frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12}+\ldots$ is equal to

If $\alpha = \frac{5}{2! \times 3} + \frac{5 \times 7}{3! \times 3^2} + \frac{5 \times 7 \times 9}{4! \times 3^3} + \ldots$,then $\alpha^2 + 4\alpha =$