Question

If p is a real number and if the middle term in the expansion of ${[\frac{P}{2}+2]}^8$ is $1120$, find p.

Answer 1

Since n = 8, therefore total terms in the

expansion is n + 1= 9 Hence it will

have only one middle term i.e $(\frac{n}{2}+1{)}^{th}term=5^{th}term$

$T_{r+1}=T_5{=}^8{C}_4(\frac{p}{2}{)}^4{2}^4=1120$

$\Rightarrow \frac{8!}{4!4!}\frac{p^4}{2^4}=1120$

$p^4=\frac{1120\times 4\times 3\times 2\times 4!}{8\times 7\times 6\times 5\times 4!}=\frac{1120}{70}=16$

$\Rightarrow P=\pm 2$