QJEEM23Ma1d59: JEE Main 2023 Mathematics - Class 11 JEEM23Ma1d59 Solution: The absolute difference of the coefficients of and in the…

Question

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

Now,

22-2 r=10

and

22-3 r=7

\begin{array}{ll}\Rightarrow 3 r=12 &amp;&amp;&amp; \Rightarrow 3 r=15 \\\\\Rightarrow r=4 &amp;&amp;&amp; \Rightarrow r=5\end{array}

\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}&amp; ={ }^{11} \mathrm{C}_4 \times 8-{ }^{11} \mathrm{C}_5 \times 2 \\\\&amp; =\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}&amp; =\frac{11 \times 10 \times 9 \times 8 \times 8}{24}-\frac{11 \times 10 \times 9 \times 8 \times 7 \times 2}{120} \\\\&amp; =11 \times 10 \times 8 \times 3-11 \times 3 \times 4 \times 7 \\\\&amp; =11 \times 3 \times 4[20-7] \\\\&amp; =11 \times 12 \times 13=(12-1) \times 12 \times(12+1) \\\\&amp; =12\left(12^2-1\right)=12^3-12\end{aligned}

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