Question

The sum of the last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ is
A. ${2^{16}}$
B. ${2^{15}}$
C. ${2^{14}}$
D. None of these

Answer 1

Hint: This problem is based on the Binomial Theorem that helps to expand the expression in the form of ${(x + y)^n}$ hence, to determine the last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ , we have to determine the value of sum of binomial coefficients each starting from $8$ and ending at $15$ .

Formula used:
The formula of binomial expansion of ${\left( {1 + x} \right)^n}$ is used in this problem which is defined as: -
${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1} \cdot {x^1} + {}^n{C_2} \cdot {x^2} + ........... + {}^n{C_n} \cdot {x^n}$

Complete Step by Step Solution:
The last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ must be ${}^{15}{C_8} + {}^{15}{C_9} + {}^{15}{C_{10}} + ........... + {}^{15}{C_{15}}$.
Now, using formula of expansion of ${\left( {1 + x} \right)^n} = {}^n{C_0} + {}^n{C_1} \cdot {x^1} + {}^n{C_2} \cdot {x^2} + ........... + {}^n{C_n} \cdot {x^n}$ , we get
The binomial expansion of ${\left( {1 + x} \right)^{15}}$ will be:
${\left( {1 + x} \right)^{15}} = {}^{15}{C_0} + {}^{15}{C_1} \cdot {x^1} + {}^{15}{C_2} \cdot {x^2} + ........... + {}^{15}{C_{15}} \cdot {x^{15}}$
Now, to find the sum of coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ , substitute $x = 1$
${\left( {1 + 1} \right)^{15}} = {}^{15}{C_0} + {}^{15}{C_1} \cdot {\left( 1 \right)^1} + {}^{15}{C_2} \cdot {\left( 1 \right)^2} + ........... + {}^{15}{C_{15}} \cdot {\left( 1 \right)^{15}}$
$ \Rightarrow {2^{15}} = {}^{15}{C_0} + {}^{15}{C_1} + {}^{15}{C_2} + ........... + {}^{15}{C_{15}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
But from permutation and combination we know that ${}^n{C_r} = {}^n{C_{n - r}}$
i.e., ${}^{15}{C_{15}} = {}^{15}{C_0},{}^{15}{C_{10}} = {}^{15}{C_5},{}^{15}{C_9} = {}^{15}{C_6}\,and\,so\,on....$
Now, from equation $(1)$
$ \Rightarrow {2^{15}} = {}^{15}{C_{15}} + {}^{15}{C_{14}} + {}^{15}{C_{13}} + ..... + {}^{15}{C_8} + {}^{15}{C_8} + {}^{15}{C_9} + {}^{15}{C_{10}} + ....... + {}^{15}{C_{15}}$
$ \Rightarrow 2\left( {{}^{15}{C_8} + {}^{15}{C_9} + {}^{15}{C_{10}} + ........... + {}^{15}{C_{15}}} \right) = {2^{15}}$
On further calculation, we get
$ \Rightarrow {}^{15}{C_8} + {}^{15}{C_9} + {}^{15}{C_{10}} + ........... + {}^{15}{C_{15}} = \dfrac{{{2^{15}}}}{2} = {2^{14}}$
Thus, the sum of the last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$ is ${2^{14}}$.
Hence, the correct option is (C) ${2^{14}}$.

Note: Before solving this question, we must analyze the given data very carefully, as it is just a formula-based question, therefore, we have to use a simple binomial expansion formula only to determine the sum of last eight coefficients in the expansion of ${\left( {1 + x} \right)^{15}}$.