If P be the sum of odd terms and Q that of even terms in the expansion of ( x - Maths - Binomial theorem

Question

If P be the sum of odd terms and Q that of even terms in the
expansion of ( x + a )^n prove that 2(P^2+Q^2) = [(x+a)^2n -
(x-a)^2n]

Ajanta Trivedi · Accepted Answer

(x+a)n=xn+C1n xn-1 a+C2n xn-2 a2+C3n xn-3 a3+ +an=(xn+C2n xn-2 a2+
)+(C1n xn-1 a+C3n xn-3 a3+ )=P+Q   (1)
(x-a)n=xn-C1n xn-1 a+C2n xn-2 a2-C3n xn-3 a3+ +(-1)nan=(xn+C2n xn-2
a2+ )-(C1n xn-1 a+C3n xn-3 a3+ )=P-Q   (2)

since 2(P2+Q2)=(P+Q)2+(P-Q)2
2 (P2+Q2)=[(x+a)n]2+[(x-a)n]2=(x+a)2n+(x-a)2n

please check your question again there must be plus sign in RHS

hope this helps you

If P be the sum of odd terms and Q that of even terms in the expansion of ( x + a )^n prove that 2(P^2+Q^2) = [(x+a)^2n - (x-a)^2n]