Questions
If A is the sum of the odd terms and B the sum of even terms in the expansion of (x + a)n, then
a
(x2+a2)n
b
(x2−a2)n
c
2(x2−a2)n
d
none of these
detailed solution
Correct option is B
We have, (x+a)n=nC6xn+nC1xn−1a1+nC2xn−2a2+nC3xn−3a3+…+nCnxn =(nC0xn+nC2xn−2a2+…)+(nC1xn−1a1+nC3xn−3a3+…) =A+B (x−a)n=nC0xn−nC1xn−1a1+nC2xn−2a2−nC3xn−3a3+…+nCn(−1)nan =(nC0xn+nC2xn−2a2+…)−(nC1xn−1a1+nC3xn−3a3+…) =A−B ∴ A2−B2=(A+B)(A−B)=(x+a)n(x−a)n =(x2−a2)nTalk to our academic expert!
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