Binomial Theorem And Its Simple Applications Question 102
Question: In the expansion of $ {{(x+a)}^{n}} $ , the sum of odd terms is P and sum of even terms is Q, then the value of $ (P^{2}-Q^{2}) $ will be
[RPET 1997; Pb. CET 1998]
Options:
A) $ {{(x^{2}+a^{2})}^{n}} $
B) $ {{(x^{2}-a^{2})}^{n}} $
C) $ {{(x-a)}^{2n}} $
D) $ {{(x+a)}^{2n}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {{(x+a)}^{n}}=x^{n}+{{,}^{n}}C_1{x^{n-1}}a{{+}^{{}}} $ ….. $ =(x^{n}+{{,}^{n}}C_2{x^{n-2}}a^{2}+……. $ + $ {{(}^{n}}C_1{x^{n-1}}a+{{,}^{n}}C_3{x^{n-3}}a^{3}+…..) $ = $ P+Q $
$ \therefore $ $ {{(x-a)}^{n}}=P-Q, $ As the terms are alter. +ve and -ve
$ \therefore $ $ P^{2}-Q^{2}=(P+Q)(P-Q)={{(x+a)}^{n}}{{(x-a)}^{n}} $
$ P^{2}-Q^{2}={{(x^{2}-a^{2})}^{n}} $
BETA
