Question

If the sum of the coefficients in the expansion of ${{(a+b)}^{n}}=4096$ , then the greatest coefficient in the expansion is
(a) 924
(b) 792
(c) 1594
(d) None of these

Answer 1

Use binomial expansion of ${{(a+b)}^{n}}$ where $n$ is a positive integer and $a,b$ are real numbers then ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are known as binomial coefficients. Put $a=1$ and $b=1$ in the given expansion. The greatest coefficient in the expansion will be the value of the middle coefficient.

Complete step by step answer:
In the question it is given that the sum of the coefficients in the expansion of ${{(a+b)}^{n}}=4096$ . Now we will expand ${{(a+b)}^{n}}$ using binomial expansion ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where $n$ is a positive integer and $a,b$ are real number and ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are binomial coefficients.
Putting $a=1$ and $b=1$ in the given expansion of ${{(a+b)}^{n}}$ we get,
$\begin{aligned} & {{(1+1)}^{n}}={}^{n}{{C}_{0}}{{(1)}^{n}}+{}^{n}{{C}_{1}}{{(1)}^{n-1}}(1)+{}^{n}{{C}_{2}}{{(1)}^{n-2}}{{(1)}^{2}}+...+{}^{n}{{C}_{n-1}}(1){{(1)}^{n-1}}+{}^{n}{{C}_{n}}{{(1)}^{n}} \\\ & {{(2)}^{n}}={}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+...+{}^{n}{{C}_{n-1}}+{}^{n}{{C}_{n}} \\\ \end{aligned}$
We are given that sum of coefficients ${}^{n}{{C}_{0}}+{}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}+...+{}^{n}{{C}_{n-1}}+{}^{n}{{C}_{n}}$ is equal to 4096. So the above equation now becomes,
${{(2)}^{n}}=4096$
We will try to write 4096 as the power of 2 so that we can get the value of $n$ by comparing both sides of the equation.
${{(2)}^{n}}={{(2)}^{12}}$
So the value of $n=12,$ which is an even number. This value of n implies there are 13 terms in total in the expansion of ${{(a+b)}^{12}}$ . In 13 terms the middle term’s coefficient has the greatest value. That is
${}^{n}{{C}_{\dfrac{n}{2}}}={}^{12}{{C}_{\dfrac{12}{2}}}={}^{12}{{C}_{6}}$
Thus coefficient ${}^{12}{{C}_{6}}$ is the greatest value. Solving ${}^{12}{{C}_{6}}$ we get,
$\begin{aligned} & {}^{12}{{C}_{6}}=\dfrac{12!}{6!(12-6)!} \\\ & =\dfrac{12!}{(6!)(6!)} \\\ \end{aligned}$
After solving the factorial value we get the value of ${}^{12}{{C}_{6}}$ as,
$\begin{aligned} & {}^{12}{{C}_{6}}=\dfrac{12\times 11\times 10\times 9\times 8\times 7}{6\times 5\times 4\times 3\times 2\times 1} \\\ & =924 \\\ \end{aligned}$
Therefore the value of ${}^{12}{{C}_{6}}=924$ which is the greatest coefficient value in the expansion of ${{(a+b)}^{n}}$ .

So, the correct answer is “Option A”.

Note: Binomial expansion of ${{(a+b)}^{n}}$ where $n$ is a positive integer and $a,b$ are real numbers then ${{(a+b)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}+{}^{n}{{C}_{1}}{{a}^{n-1}}b+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}+...+{}^{n}{{C}_{n-1}}a{{b}^{n-1}}+{}^{n}{{C}_{n}}{{b}^{n}}$ where ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},{}^{n}{{C}_{2}},...,{}^{n}{{C}_{n}}$ are known as binomial coefficients. If we have to expand ${{(a+b)}^{n}}$ in the binomial expansion then the middle terms coefficient will be the greatest value that is if n is even then ${}^{n}{{C}_{\dfrac{n}{2}}}$ will give the greatest value.