# Question 6 Exercise 7.2

Solutions of Question 6 of Exercise 7.2 of Unit 07: Permutation, Combination and Probablity. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

## Question 6

Find the constant term in the expansion or $(2 \sqrt{x}-\dfrac{3}{x \sqrt{x}})^{23}$

### Solution

Since we see that $a=2 \sqrt{x}$, $b=-\dfrac{3}{x \sqrt{x}}$ and $n=23$.

By constant term, we mean the term that is independent of $x$.

The general term of the given expansion is: \begin{align} T_{r+1}&=\dfrac{23 !}{(23-r) ! r !}(2 \sqrt{x})^{23-r}(-\dfrac{3}{x \sqrt{x}})^r \\ & =\dfrac{23 !}{(23-r) ! r !} \cdot 2^{23-r} \cdot(-3)^r \cdot x^{\dfrac{23-r}{2}} \cdot(x^{-\dfrac{3}{2}})^r\\ & =\dfrac{23 !}{(23-r) ! r !} \cdot 2^{23-r} \cdot(-3)^r x \dfrac{23-r}{2}-\dfrac{3 r}{2} \\ & =\dfrac{23 !}{(23-r) ! r !} \cdot 2^{23-r} \cdot(-3)^r \cdot x \dfrac{23-4 r}{2}\end{align} But the term $T_{r+1}$ is independent of $x$ is possible only is \begin{align}x^{\dfrac{23-4 r}{2}}&=x^0\\ \Rightarrow 23-4 r&=0 \\ \Rightarrow r&=\dfrac{23}{4}\end{align} Which is not possible because $r$ should be a positive integer.

It means there is no constant term or term independent of $x$ in the expansion of $(2 \sqrt{x}-\dfrac{3}{x \sqrt{x}})^{23}$.

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