Binomial Theorem for Negative and Fractional Indices

If ${\left(1-x\right)}^{-n}=a_0+a_{1}x+a_2{x}^2+\dots ..+a_r{x}^r+\dots$, then $a_0+a_1+a_2+\dots +a_r$ is equal to

If $-\frac{1}{3}<x<1.$ Then
$1+n\left(\frac{2x}{1+x}\right)+\frac{n(n+1)}{2!}{\left(\frac{2x}{1+x}\right)}^2+\dots \dots \dots \dots$
is equal to

If $x=\frac{5}{2!3}+\frac{5·7}{3!3^2}+\frac{5·7·9}{4!3^3}+\dots ,$ then $x^2+4x=$

What is the coefficient of $\frac{y^3}{x^8}$ in $(x+y{)}^{-5},$ when $\left|\frac{y}{x}\right|<1?$

The sum of the series $1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2\times 5}{3\times 6} {\left(\frac{1}{8}\right)}^2+\frac{2\times 5\times 8}{3\times 6\times 9} {\left(\frac{1}{8}\right)}^3+\dots$

The interval in which the expansion of $\frac{3x}{\left(x-2\right) \left(x-\mathrm{1 }\right)}$is valid

The coefficient of $x^n$ in the expression of $\frac{1}{x^2-5x+6} for \left|x\right|<1$ is

If $\left|x\right|$ is so small so that $x^2$ and higher powers of x may be neglected, then an approximate value of $\frac{{\left(1+\frac{2}{3}x\right)}^{-3}{\left(1-15x\right)}^{\frac{-1}{5}} }{{\left(2-3x\right)}^4}$ is

If $x=1+\frac{3}{1!}\times \frac{1}{6}+\frac{3\times 7}{2!}{\left(\frac{1}{6}\right)}^2+\frac{3\times 7\times 11}{3!}{\left(\frac{1}{6}\right)}^3 +\dots \infty$, then $x^4$ is

If $x=\frac{1}{5}+\frac{1.3}{5.10}+\frac{\mathrm{1.3.5}}{\mathrm{5.10.15}}+\dots \infty ,$ then $3x^2+6x=$

If $\left|x\right|<1$ then the coefficient of $x^5$ in the expansion of $\frac{3x}{\left(x-2\right) \left(x+1\right)}$ is

If $\left|x\right|<1$ then the coefficient of $x^5$ in the expansion of $\frac{3x}{\left(x-2\right) \left(x+1\right)}$ is

If ${\left(\alpha +bx\right)}^{-3}=\frac{1}{27}+\frac{1}{3}x+\dots .,$ then the ordered pair (a, b) equals to

If $x={\left(2+\sqrt{3}\right)}^n$ , then the value of $x-x^2+x\left[x\right], where \left[\cdot \right]$ denotes the greatest integer function, is equal to

The value of $\frac{2\frac{1}{2} }{1!}+\frac{3\frac{1}{2}}{2!}+\frac{4\frac{1}{2}}{3!}+\frac{5\frac{1}{2}}{4!}+\dots \infty$ is

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8\left(2\right)!}+\frac{1}{16\left(3\right)!}+\frac{1}{32\left(4\right)!}+\dots \infty =$

If $y=-\left(x^3+\frac{x^6}{2}+\frac{x^9}{3}+\dots \right), then x=$

If $y=x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\dots then x=$

If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \infty ,$ then $x=$

If the ratio of the coefficient of third and fourth term in the expansion of ${\left(x-\frac{1}{2x}\right)}^n$is $1:2$, then the value of $n$ will be