The term independent of x in the binomial expansion of (1-1x+3x^5)(2x^2-1x)^8 is:

Question

The term independent of $x$ in the binomial expansion of $\left(1-\dfrac1x+3x^5\right)\left(2x^2-\dfrac1x\right)^8$ is:

Accepted Answer

General term of $\left(2x^2-\dfrac1xight)^8$: $T_{r+1}=\binom8r2^{8-r}(-1)^rx^{16-3r}$

Multiplying by $1,\ -\dfrac1x,\ 3x^5$ requires powers $16-3r\in\{0,1,-5\}$:

- $16-3r=0$: not integer
- $16-3r=1$: $r=5$: contributes $-\binom85 2^3(-1)^5=-(-448)=448$... 
- $16-3r=-5$: $r=7$: contributes $3	imes\binom87 2(-1)^7=3	imes(-16)=-48$

Following the source's exact computation: total $=(56	imes8)-48=448-48=400$

Solutions · 1