The absolute difference of the coefficients of x^10 and x^7 in the expansion of (2x^2+12x)…

Question

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $\left(2x^2+\dfrac{1}{2x}\right)^{11}$ is equal to

Accepted Answer

$$T_{r+1}=\binom{11}{r}(2x^2)^{11-r}\left(\frac{1}{2x}\right)^r=\binom{11}{r}2^{11-2r}x^{22-3r}$$

For $x^{10}$: $r=4$; for $x^7$: $r=5$.

$$\text{Coeff of }x^{10}=\binom{11}{4}2^{3},\qquad\text{Coeff of }x^7=\binom{11}{5}2^{1}$$

$$\text{Difference}=\binom{11}{4}\times8-\binom{11}{5}\times2 = 11\times10\times3\times8-11\times3\times4\times7=11\times3\times4\times(20-7)=11\times12\times13$$

$=12(12^2-1)=12^3-12$

Solutions · 1