First slide

Binomial theorem for positive integral Index

Question

The middle term in the expansion of $(x / 2 + 2)^{8}$ is 1120; then $x \in R$ is equal to

Moderate

A

-2
B

3
C

-3
D

2

Solution

Middle term is $(\frac{n}{2} + 1) th or (4 + 1)^{th} or T_{5}$
$\begin{array}{l} \Rightarrow T_{5} =^{8} C_{4} {(\frac{x}{2})}^{4} \times 2^{4} = 1120 \\ Or \frac{8 \times 7 \times 6 \times 5}{1 \times 2 \times 3 \times 4} x^{4} = 1120 \\ or x^{4} = \frac{1120}{70} = 16 \\ or (x^{2} + 4) (x^{2} - 4) = 0 \\ or x = \pm 2 \end{array}$

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