If P be the sum of odd terms and Q be the sum of even terms in the expansion of - Maths - Binomial Theorem

Question

If P be the sum of odd terms and Q be the sum of even terms in the
expansion of (x + a) to the power of n, prove that 4 PQ = ( x + a)
to the power of 2n - ( x - a) to the power of 2n?

Varun.Rawat · Accepted Answer

Solution :

(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)Squaring 1 and 2 and then subtracting, we get(x+a)2n-(x-a)2n = P+Q2 - P-Q2 = P2+Q2+2PQ-P2-Q2+2PQ = 4PQ

If P be the sum of odd terms and Q be the sum of even terms in the expansion of (x + a) to the power of n, prove that 4 PQ = ( x + a) to the power of 2n - ( x - a) to the power of 2n?