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

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(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 $........$.

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