The coefficient of x^7 in the expansion of (1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.

Vedclass · Accepted Answer

$-144$

Vedclass · Answer

(B) Given expression: $(1 - x - x^2 + x^3)^6 = ((1 - x)(1 - x^2))^6 = (1 - x)^6 (1 - x^2)^6 = (1 - x)^6 (1 - x)^6 (1 + x)^6 = (1 - x)^{12} (1 + x)^6$.We need the coefficient of $x^7$ in $(1 - x)^{12} (1 + x)^6$.Let $(1 - x)^{12} = \sum_{i=0}^{12} (-1)^i \binom{12}{i} x^i$ and $(1 + x)^6 = \sum_{j=0}^{6} \binom{6}{j} x^j$.The coefficient of $x^7$ is $\sum_{i+j=7} (-1)^i \binom{12}{i} \binom{6}{j}$.For $i=1, j=6: -\binom{12}{1} \binom{6}{6} = -12 	imes 1 = -12$.For $i=2, j=5: \binom{12}{2} \binom{6}{5} = 66 	imes 6 = 396$.For $i=3, j=4: -\binom{12}{3} \binom{6}{4} = -220 	imes 15 = -3300$.For $i=4, j=3: \binom{12}{4} \binom{6}{3} = 495 	imes 20 = 9900$.For $i=5, j=2: -\binom{12}{5} \binom{6}{2} = -792 	imes 15 = -11880$.For $i=6, j=1: \binom{12}{6} \binom{6}{1} = 924 	imes 6 = 5544$.For $i=7, j=0: -\binom{12}{7} \binom{6}{0} = -792 	imes 1 = -792$.Summing these: $-12 + 396 - 3300 + 9900 - 11880 + 5544 - 792 = -144$.

The coefficient of $x^7$ in the expansion of $(1 - x - x^2 + x^3)^6$ is

Similar Questions

Let $\alpha$ be the constant term in the binomial expansion of $\left(\sqrt{x} - \frac{6}{x^{3/2}}\right)^n$,$n \leq 15$. If the sum of the coefficients of the remaining terms in the expansion is $649$ and the coefficient of $x^{-n}$ is $\lambda \alpha$,then $\lambda$ is equal to $..........$.

If the sum of the coefficients in the expansion of $(x - 2y + 3z)^n$,$n \in N$ is $128$,then the greatest coefficient in the expansion of $(1 + x)^n$ is

For each positive integer $n$,let $A_n = \max \left\{ \binom{n}{r} \mid 0 \leq r \leq n \right\}$. Then,the number of elements $n \in \{1, 2, \ldots, 20\}$ for which $1.9 \leq \frac{A_n}{A_{n-1}} \leq 2$ is

The number of integral terms in the expansion of $(\sqrt{3} + \sqrt[8]{5})^{256}$ is

If the coefficient of $x^7$ in $(ax - \frac{1}{bx^2})^{13}$ and the coefficient of $x^{-5}$ in $(ax + \frac{1}{bx^2})^{13}$ are equal,then $a^4 b^4$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module