Binomial Theorem for Rational Index

$1+\frac{1}{3}x+\frac{1\times 4}{3\times 6}{x}^2+\frac{1\times 4\times 7}{3\times 6\times 9}{x}^3+\dots$ is equal to

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 $\left|\frac{y}{x}\right|<1?$

If $x=(\sqrt{3}+1{)}^{2n+1},$ then $\left[x\right]$ is [Note : where $\left[x\right]$ is greatest integer function.]

The numeric value of coefficient of $t^4$ in the expansion of the term ${\left(\frac{1-t^6}{1-t}\right)}^3$ is:

The value of $1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3.5}{3.6.9}+\dots \infty , \mathrm{i}\mathrm{s}$

Find the coefficient of $x^4$ in the expansion of ${\left(1-9x+20x^2\right)}^{-1}$.

If $0<\left|x\right|<1,$ the coefficient of $x^n$ in the expansion of ${\left(1-x\right)}^{-\text{1}}$is _____

For $\left|x\right|<1,$ the coefficient of the term independent of $x$ in the expansion of $\frac{1}{{\left(x-1\right)}^2(x-2)}$is______.

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

The coefficient of $x$ in the expansion of ${\left(1+x+x^2+x^3\right)}^{-3}$ is

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