# Problem Study : The Binomial Theorem

It is given that the coefficient of $\displaystyle {{x}^{2}}$ is equal to the coefficient of  $\displaystyle {{x}^{3}}$ in the binomial expansion of $\displaystyle {{(2k\text{ }+\text{ }x)}^{n}}$, where $\displaystyle k$ is a constant and $\displaystyle n$ is a positive integer. Prove that $\displaystyle n=6k+2$.

Solution

$\displaystyle {{(r+1)}^{{th}}}$ term in the expansion of $\displaystyle {{(2k\text{ }+\text{ }x)}^{n}}={}^{n}{{C}_{r}}{{(2k)}^{{n-r}}}{{x}^{r}}$

For  $\displaystyle {{x}^{2}}$, $\displaystyle r=2$

$\displaystyle \therefore$ coefficient of $\displaystyle {{x}^{2}}={}^{n}{{C}_{2}}{{(2k)}^{{n-2}}}$

For  $\displaystyle {{x}^{3}}$, $\displaystyle r=3$

$\displaystyle \therefore$ coefficient of $\displaystyle {{x}^{3}}={}^{n}{{C}_{3}}{{(2k)}^{{n-3}}}$

By the problem,

Coefficient of $\displaystyle {{x}^{2}}=$ Coefficient of $\displaystyle {{x}^{3}}$

$\displaystyle \therefore {}^{n}{{C}_{2}}{{(2k)}^{{n-2}}}={}^{n}{{C}_{3}}{{(2k)}^{{n-3}}}$

$\displaystyle \therefore \frac{{n(n-1)}}{{1\times 2}}(2k)=\frac{{n(n-1)(n-2)}}{{1\times 2\times 3}}$

$\displaystyle \therefore 2k=\frac{{n-2}}{3}$

$\displaystyle \therefore n=6k+2$