Question

The coefficient of ${x^{65}}$ in the expansion of ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$ is
(A) ${}^{130}{c_{65}} + {}^{129}{c_{66}}$
(B) ${}^{130}{c_{65}} + {}^{129}{c_{55}}$
(C) ${}^{130}{c_{66}} + {}^{129}{c_{65}}$
(D) none of these.

Answer 1

Hint: We have given the expression ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$ , to find the coefficient of so will simplify by splitting the first term to make the exponent the same as ${a^n}{b^n} = {\left( {ab} \right)^n}$. Then we will get a expression as a product of two factors so we will simplify it such a way that it will split as the sum of terms then we will find the coefficient of ${x^{65}}$ from the terms separately using the formula
  \[{(1 + x)^n} = 1 + {}^n{c_1}x + {}^n{c_2}{x^2} + {}^n{c_3}{x^3} + \cdots + {}^n{c_k}{x^k} + \cdots + {}^n{c_n}{x^n}\].Now the summation of the coefficients found from those two terms will be the required coefficient.

Complete step by step answer:

Given ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$
 ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$
On splitting the first term we get,
 $ = (1 + x) \times {(1 + x)^{130}} \times {({x^2} - x + 1)^{130}}$
Taking the last two terms together, we get,
 $ = (1 + x) \times {\left[ {(1 + x)({x^2} - x + 1)} \right]^{130}}$
Since ${a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})$ , we get,
 $ = (1 + x) \times {\left[ {1 + {x^3}} \right]^{130}}$
On multiplication we get,
 $ = {\left( {1 + {x^3}} \right)^{130}} + x{\left( {1 + {x^3}} \right)^{130}}$

Now, we have to find the coefficient of ${x^{65}}$ in the expansion of ${\left( {1 + {x^3}} \right)^{130}} + x{\left( {1 + {x^3}} \right)^{130}}$ ,
i.e., the coefficient of ${x^{65}}$ in the expansion of ${\left( {1 + {x^3}} \right)^{130}}$ + coefficient of ${x^{65}}$ in the expansion of $x{\left( {1 + {x^3}} \right)^{130}}$
The expansion of ${\left( {1 + {x^3}} \right)^{130}}$ is given by,
 ${\left( {1 + {x^3}} \right)^{130}} = 1 + {}^{130}{c_1}\left( {{x^3}} \right) + {}^{130}{c_2}{\left( {{x^3}} \right)^2} + {}^{130}{c_3}{\left( {{x^3}} \right)^3} + \cdots + {}^{130}{c_k}{x^{3k}} + \cdots + {}^{130}{c_n}{x^{3n}}$
Note that, here we have only those powers of x that are multiples of 3.
Since 65 is not a multiple of 3, therefore coefficient of ${x^{65}}$ in the expansion of ${\left( {1 + {x^3}} \right)^{130}}$ is 0.

Again, in order to find the coefficient of ${x^{65}}$ in the expansion of $x{\left( {1 + {x^3}} \right)^{130}}$, we have to find the coefficient of ${x^{64}}$ in the expansion of${\left( {1 + {x^3}} \right)^{130}}$.
64 is again not a multiple of 3. Therefore the coefficient of ${x^{64}}$ in the expansion of${\left( {1 + {x^3}} \right)^{130}}$is 0.

Therefore, the coefficient of ${x^{65}}$ in the expansion of${\left( {1 + {x^3}} \right)^{130}} + x{\left( {1 + {x^3}} \right)^{130}}$ is 0
This implies that the coefficient of ${x^{65}}$ the expansion of ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$ is 0.
Hence, (D) is the correct option.

Note: Most of the students try to find the coefficient of ${x^{65}}$ from the expression ${(1 + x)^{131}}{({x^2} - x + 1)^{130}}$ like such a way that finding the coefficient of ${x^n}$from the first factor and the coefficient of ${x^{65 - n}}$ from the second factor.
Using this we will get a series to solve for the coefficient of ${x^{65}}$, so we will have to solve that which will make the problem much more complex, so we should try to avoid that solution and adapt for the above solution to get the required answer more simply and properly.
Note that, the expansion of ${(1 + x)^n}$ is given by:
\[{(1 + x)^n} = 1 + {}^n{c_1}x + {}^n{c_2}{x^2} + {}^n{c_3}{x^3} + \cdots + {}^n{c_k}{x^k} + \cdots + {}^n{c_n}{x^n}\]
Therefore, the coefficient of \[{x^k}\] in the expansion of ${(1 + x)^n}$ is\[{}^n{c_k}{x^k}\]