The term independent of x in the binomial exp | vedclass

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Question

(C) The general term of $\left( 2x^2 - \frac{1}{x} \right)^8$ is given by $T_{r+1} = ^8C_r (2x^2)^{8-r} (-x^{-1})^r = ^8C_r 2^{8-r} (-1)^r x^{16-3r}$.

Vedclass · Accepted Answer

$400$

Vedclass · Answer

(C) The general term of $\left( 2x^2 - \frac{1}{x} \right)^8$ is given by $T_{r+1} = ^8C_r (2x^2)^{8-r} (-x^{-1})^r = ^8C_r 2^{8-r} (-1)^r x^{16-3r}$.
The expression is $\left( 1 - x^{-1} + 3x^5 \right) \sum_{r=0}^8 {^8C_r} 2^{8-r} (-1)^r x^{16-3r}$.
Expanding this,we get:
$1 \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r} - x^{-1} \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r} + 3x^5 \cdot \sum ^8C_r 2^{8-r} (-1)^r x^{16-3r}$.
For the term independent of $x$:
$1$. From the first part,$16-3r = 0 \implies r = 16/3$ (not an integer).
$2$. From the second part,$16-3r-1 = 0 \implies 15-3r = 0 \implies r = 5$. The coefficient is $- ^8C_5 2^{8-5} (-1)^5 = -56 \cdot 8 \cdot (-1) = 448$.
$3$. From the third part,$16-3r+5 = 0 \implies 21-3r = 0 \implies r = 7$. The coefficient is $3 \cdot ^8C_7 2^{8-7} (-1)^7 = 3 \cdot 8 \cdot 2 \cdot (-1) = -48$.
Summing these,the constant term is $448 - 48 = 400$.

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

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