If $y = 3x + 6x^2 + 10x^3 + \dots$,then the value of $x$ in terms of $y$ is

Assertion $(A)$: If $|x| < 1$,then $\sum_{n=0}^{\infty}(-1)^n x^{n+1} = \frac{x}{x+1}$.
Reason $(R)$: If $|x| < 1$,then $(1+x)^{-1} = 1-x+x^2-x^3+\dots$.
Which one of the following is true?

If $x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots$ to infinite terms,then $9x^2 + 24x = $

If $|x| < 1$,then the value of $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$ will be

If $x$ is a positive real number and the first negative term in the expansion of $(1+x)^{27/5}$ is $t_k$,then $k=$

Assertion $(A) : 1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8}+\ldots \infty = \sqrt[3]{4}$
Reason $(R) : |x| < 1, (1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$ The correct answer is