Find the coefficient of x^5 in (1+x+x^2)^8. | vedclass

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

Question

(D) The general term in the expansion of $(1+x+x^2)^8$ is given by $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (x^2)^{n_3}$,where $n_1 + n_2 + n_3

Vedclass · Accepted Answer

$504$

Vedclass · Answer

(D) The general term in the expansion of $(1+x+x^2)^8$ is given by $\frac{8!}{n_1! n_2! n_3!} (1)^{n_1} (x)^{n_2} (x^2)^{n_3}$,where $n_1 + n_2 + n_3 = 8$ and $n_2 + 2n_3 = 5$.We find the non-negative integer solutions for $(n_1, n_2, n_3)$:$1$. If $n_3 = 0$,then $n_2 = 5$,so $n_1 = 8 - 5 - 0 = 3$. Coefficient: $\frac{8!}{3! 5! 0!} = 56$.$2$. If $n_3 = 1$,then $n_2 = 3$,so $n_1 = 8 - 3 - 1 = 4$. Coefficient: $\frac{8!}{4! 3! 1!} = 280$.$3$. If $n_3 = 2$,then $n_2 = 1$,so $n_1 = 8 - 1 - 2 = 5$. Coefficient: $\frac{8!}{5! 1! 2!} = 168$.Summing these coefficients: $56 + 280 + 168 = 504$.Thus,the coefficient of $x^5$ is $504$.

Find the coefficient of $x^5$ in $(1+x+x^2)^8$.

Similar Questions

The coefficient of $x$ in the expansion of $[\sqrt{1 + x^2} - x]^{-1}$ in ascending powers of $x$,when $|x| < 1$,is

The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3) \ldots (1+x^{100})$ is

If $f(x) = x^n$,then the value of $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!}$ is

The sum of the coefficients in the expansion of $(1+x+x^2)^n$ is

Let $[x]$ denote the greatest integer less than or equal to $x$. If for $n \in N$,$(1-x+x^3)^n = \sum_{j=0}^{3n} a_j x^j$,then $\sum_{j=0}^{[\frac{3n}{2}]} a_{2j} + 4 \sum_{j=0}^{[\frac{3n-1}{2}]} a_{2j+1}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module