In the expansion of (1 + 3x + 2x^2)^6,the coe | vedclass

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

Question

(D) Given the expression $(1 + 3x + 2x^2)^6 = ((1 + x)(1 + 2x))^6 = (1 + x)^6 (1 + 2x)^6$.We need the coefficient of $x^{11}$ in the product $(1 + x)^

Vedclass · Accepted Answer

$576$

Vedclass · Answer

(D) Given the expression $(1 + 3x + 2x^2)^6 = ((1 + x)(1 + 2x))^6 = (1 + x)^6 (1 + 2x)^6$.We need the coefficient of $x^{11}$ in the product $(1 + x)^6 (1 + 2x)^6$.$(1 + x)^6 = \sum_{r=0}^{6} \binom{6}{r} x^r$ and $(1 + 2x)^6 = \sum_{k=0}^{6} \binom{6}{k} (2x)^k = \sum_{k=0}^{6} \binom{6}{k} 2^k x^k$.The general term in the product is $\binom{6}{r} x^r \cdot \binom{6}{k} 2^k x^k = \binom{6}{r} \binom{6}{k} 2^k x^{r+k}$.For $x^{11}$,we need $r + k = 11$. Since $0 \le r, k \le 6$,the possible pairs $(r, k)$ are $(5, 6)$ and $(6, 5)$.For $(r, k) = (5, 6)$: Coefficient is $\binom{6}{5} \binom{6}{6} 2^6 = 6 \cdot 1 \cdot 64 = 384$.For $(r, k) = (6, 5)$: Coefficient is $\binom{6}{6} \binom{6}{5} 2^5 = 1 \cdot 6 \cdot 32 = 192$.Total coefficient of $x^{11} = 384 + 192 = 576$.

In the expansion of $(1 + 3x + 2x^2)^6$,the coefficient of $x^{11}$ is

Similar Questions

If the coefficients of the $(2r + 4)^{th}$ and $(r - 2)^{th}$ terms in the expansion of $(1 + x)^{18}$ are equal,then $r =$

The coefficient of $x^4$ in the expansion of $\frac{(1-3 x)^2}{(1-2 x)}$ is equal to

If the $9^{th}$ and $10^{th}$ terms are the numerically greatest terms in the expansion of $(5x - 6y)^n$ when $x = 2/5$ and $y = 1/2$,then the absolute value of the middle term of that expansion is:

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

The sum of the rational terms in the expansion of $(\sqrt{2}+\sqrt[5]{3})^{10}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module