The absolute difference of the coefficients of x^10 and x^7 in the expansion of [(2 x^2+(1…

Question

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

Accepted Answer

General term of $\left(2 x^2+\frac{1}{2 x}\right)^{11}$ is :

$$
\begin{aligned}
& \mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_r\left(2 x^2\right){ }^{11-r}\left(\frac{1}{2 x}\right)^r \\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-2 r} 2^{-r} x^{-r} \\
& ={ }^{11} \mathrm{C}_r 2^{11-r} x^{22-3 r}
\end{aligned}
$$

Now, $22-2 r=10$ and $22-3 r=7$

$$
\[\begin{gathered} \Rightarrow 3 r=12 \quad \quad \quad  \Rightarrow 3 r=15 \ \Rightarrow r=4 \quad \quad \quad  \Rightarrow r=5 \end{gathered}$$
\]

$\therefore$ Coeff. of $x^{10}={ }^{11} \mathrm{C}_4 \cdot 2^{11-8}={ }^{11} \mathrm{C}_4 \times 8$

Coeff. of $x^7={ }^{11} C_5 \cdot 2^{11-10}={ }^{11} C_4 \times 2$

Now, required difference

$$
\begin{aligned}
& ={ }^{11} \mathrm{C}_4 \times 8-{ }^{11} \mathrm{C}_5 \times 2 \\
& =\frac{11 \times 10 \times 9 \times 8 \times 7 !}{4 ! \times 7 !} \times 8-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 ! \times 2}{5 ! 6 !}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{11 \times 10 \times 9 \times 8 \times 8}{24}-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 2}{120} \\
& =11 \times 10 \times 8 \times 3-11 \times 3 \times 4 \times 7 \\
& =11 \times 3 \times 4[20-7] \\
& =11 \times 12 \times 13=(12-1) \times 12 \times(12+1) \\
& =12\left(12^2-1\right)=12^3-12
\end{aligned}
$$

Solutions · 1