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}$

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