Question 2 Review Exercise 7

Solutions of Question 2 of Review Exercise 7 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.

Q2 Find the middle term in the expansion of $\left(2 x^3+3 y\right)^8$ ? Solution: In the above $a=2 x^3$, $b=3 y$ and $n=8$. Since $n=8$ is cven thus the number of terms are even and the middle term is $\frac{8+2}{2}=5$. Which is: $$\begin{aligned} & T_5=\frac{8 !}{(8-4) ! 4 !}\left(2 x^3\right)^{8-4}(3 y)^4 \\ & T_5=70.2^4 \cdot 3^4 \cdot x^{12} \cdot y^4 \\ & =90720 x^{12} y^4 \end{aligned}$$

Hence the middle term of the expansion is $T_5=90720 x^{12} y^4$.

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