Prove that $\sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $
By Binomial Theorem,
$\sum\limits_{r = 0}^n {{\,^n}{C_r}{a^{n - r}}{b^r} = {{\left( {a + b} \right)}^n}} $
By putting $b=3$ and $a=1$ in the above equation, we obtain
$\sum\limits_{r = 0}^n {{\,^n}{C_r}{{\left( 1 \right)}^{n - r}}{{\left( 3 \right)}^r} = {{\left( {1 + 3} \right)}^n}} $
$ \Rightarrow \sum\limits_{r = 0}^n {{3^r}{\,^n}{C_r} = {4^n}} $
Hence proved.
Find the $4^{\text {th }}$ term in the expansion of $(x-2 y)^{12}$
Let the sixth term in the binomial expansion of $\left(\sqrt{2^{\log _2}\left(10-3^x\right)}+\sqrt[5]{2^{(x-2) \log _2 3}}\right)^m$, in the increasing powers of $2^{(x-2) \log _2 3}$, be $21$ . If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an $A.P.$, then the sum of the squares of all possible values of $x$ is $.........$.
If the middle term in the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^n}$ is $924{x^6}$, then $n = $
The coefficient of ${x^{32}}$ in the expansion of ${\left( {{x^4} - \frac{1}{{{x^3}}}} \right)^{15}}$ is
The coefficient of ${x^5}$ in the expansion of ${({x^2} - x - 2)^5}$ is