Binomial Theorem for Positive Integral Indices

If $z=2+3i, \text{then} z^5+{\left(\overline{z}\right)}^5$ is equal to:

If in a series $\begin{array}{rl}& t_n=\frac{n}{(n+1)!}\end{array}$, then $\begin{array}{rl}& \sum _{\mathrm{n}=1}^{\mathrm{n}=30}{\mathrm{t}}_{\mathrm{n}}\end{array}=$

$\sum _{\mathrm{r}=0}^{\mathrm{n}}(-1{)}^{\mathrm{r}}{\mathrm{C}}_{\mathrm{r}}[\frac{1}{2^{\mathrm{r}}}+\frac{3^{\mathrm{r}}}{2^{2\mathrm{r}}}+\frac{7^{\mathrm{r}}}{2^{3\mathrm{r}}}+\frac{{15}^{\mathrm{r}}}{2^{4\mathrm{r}}}+\dots ]\text{upto}m\text{terms}$

$(x+10{)}^{50}+(x-10{)}^{50}=a_0+a_{1}x+a_2{x}^2+\dots .+a_{50}{x}^{50} for all x\in R then \frac{a_2}{a_0}=$

Expand ${\left(\sqrt{5}-2\right)}^4$ using binomial expansion.

$\frac{\left(10\times {107}^{90}+9\times {107}^{80}+8\times {107}^{70}\right)}{106}$. Remainder$= ?$

If the digit's at ten's and hundred's place in ${\left(11\right)}^{2016}$ are $x$ and $y$ respectively, then the ordered pair $\left(x,y\right)$ is equal to$?$

The sum of series $2+\frac{5}{2!3}+\frac{5·7}{3!3^2}+.....\left(\text{to infinite}\right) \text{is}$

The value of ${\left({}^{n}C_0\right)}^2+{\left({}^{n}C_1\right)}^2+{\left({}^{n}C_2\right)}^2+......+{\left({}^{n}C_n\right)}^2=$

If the coefficients of ${\left(r-5\right)}^{\mathrm{th}}$ and ${\left(2r-1\right)}^{\mathrm{th}}$ terms in the expansion of ${\left(1+x\right)}^{34}$ are equal, find $r$.

The sum of the coefficients of the first three terms in the expansion of ${\left(x-\frac{3}{x^2}\right)}^m, x\ne 0, m$ being a natural number, is $559$. Find the term of the expansion containing $x^3$.

If the coefficients of $a^{r-1}, a^r$ and $a^{r+1}$ in the expansion of ${\left(1+a\right)}^n$ are in arithmetic progression, prove that $n^2-n\left(4r+1\right)+4r^2-2=0$.

Find the term independent of $x$ in the expansion of ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^6$.

The coefficients of three consecutive terms in the expansion of ${\left(1+a\right)}^n$ are in the ratio $1: 7: 42$. Find $n$.

Using binomial theorem, prove that $6^n-5n$ always leaves remainder $1$ when divided by $25$.

Which is larger ${\left(1.01\right)}^{1000000}$ or $10,000$?

Compute ${\left(98\right)}^5$.

Expand ${\left(x^2+\frac{3}{x}\right)}^4, x\ne 0$

The sum of all the numbers in each row of Pascal’s Triangle is of the form $3^n$.

If $t=\frac{4}{5}+\frac{4\times 6}{5\times 10}+\frac{{\displaystyle 4\times 6\times 8}}{{\displaystyle 5\times 10\times 15}}+.................\infty$, then $9t=$