Binomial Theorem Question - None

Question

The term independent of x in the expression of (1 - x² + 3x³)$\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$, x $\neq$ 0 is

Accepted Answer

We are given that the expression is
(1 - x² + 3x³)$\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$; x $\neq$ 0
$\because$ General term of $\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$
¹¹C_r$\left(\frac{5}{2} x^3\right)^{11-r}\left(-\frac{1}{5 x^2}\right)^r$
[$\because$ General term of (x + y)ⁿ is ⁿC_r (x)^n-r $\cdot$ y^r]
Now general term of ¹¹C_r$\left(\frac{5}{2}\right)^{11-r}\left(-\frac{1}{5}\right)^r$x^33-5r
$\therefore$ Term independent of x is
1 $\times$ coefficient of x⁰ in $\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$ +
-1 $\times$ coefficient of x^-2 in $\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$ +
3 $\times$ coefficient of x^-3 in $\left(\frac{5}{2} x^3-\frac{1}{5 x^2}\right)^{11}$
for coefficient of x⁰
33 - 5r = 0 not possible
for coefficient of x^-2
33 - 5r = -2
$\Rightarrow$ 35 = 5r $\Rightarrow$ r = 7
for coefficient of x^-3
33 - 5r = -3
$\Rightarrow$ 36 = 5r not possible
So term independent of x is
(-1)¹¹C₇$\left(\frac{5}{2}\right)^4\left(-\frac{1}{5}\right)^7$ = $\frac{33}{200}$

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