The coefficient of x^7 in (1-x+2x^3)^{10} is | vedclass

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

Question

(A) The general term in the expansion of $(1-x+2x^3)^{10}$ is given by $\frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-x)^{r_2} (2x^3)^{r_3} = \frac{10!}{r_1!

Vedclass · Accepted Answer

$960$

Vedclass · Answer

(A) The general term in the expansion of $(1-x+2x^3)^{10}$ is given by $\frac{10!}{r_1! r_2! r_3!} (1)^{r_1} (-x)^{r_2} (2x^3)^{r_3} = \frac{10!}{r_1! r_2! r_3!} (-1)^{r_2} (2)^{r_3} x^{r_2+3r_3}$.We require $r_1+r_2+r_3=10$ and $r_2+3r_3=7$,where $r_1, r_2, r_3 \ge 0$.Possible non-negative integer solutions for $(r_1, r_2, r_3)$ are:$1$. If $r_3=0$,then $r_2=7$,so $r_1=3$.$2$. If $r_3=1$,then $r_2=4$,so $r_1=5$.$3$. If $r_3=2$,then $r_2=1$,so $r_1=7$.The coefficient is the sum of terms for these cases:Case $1$: $\frac{10!}{3! 7! 0!} (-1)^7 (2)^0 = -120$.Case $2$: $\frac{10!}{5! 4! 1!} (-1)^4 (2)^1 = 1260 	imes 2 = 2520$.Case $3$: $\frac{10!}{7! 1! 2!} (-1)^1 (2)^2 = 360 	imes (-4) = -1440$.Summing these: $-120 + 2520 - 1440 = 960$.

The coefficient of $x^7$ in $(1-x+2x^3)^{10}$ is $........$.

Similar Questions

$A$ natural number has prime factorization given by $n = 2^{x} 3^{y} 5^{z}$,where $y$ and $z$ are such that $y+z=5$ and $y^{-1}+z^{-1}=\frac{5}{6}$,with $y > z$. Then the number of odd divisors of $n$,including $1$,is ..... .

The largest $n \in \mathbb{N}$ such that $3^n$ divides $50!$ is:

If $a$ is the number of all even divisors and $b$ is the number of all odd divisors of the number $10800$,then $2a+3b=$

The sum of all positive divisors of $242$ except $1$ and itself is:

The largest natural number $n$ such that $3^{n}$ divides $66!$ is $............$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module