If P be the sum of odd terms and Q that of even terms in the expansion of (x + a)n, then the value of [(x + a)2n - (x - a)2n] equals

Question

If P be the sum of odd terms and Q that of even terms in the expansion of (x + a)n, then the value of [(x + a)2n - (x - a)2n] equals (A) PQ (B) 2PQ (C) 4 PQ (D) None of these

Tardigrade Admin · Accepted Answer

(x + a)n = (t1 + t3 + t5 + ...) + (t2 + t4 + t6 + ...) ∴ (x + a)n = P + Q ldots(i) and (x - a)n = (t1 + t3 + t5 + ...) - (t2 + t4 + t6 + ...) (x - a)n = P - Q ldots (ii) Squaring and subtracting (ii) from (i), we get (x + a)2n - (x - a)2n = (P + Q)2 - (P - Q)2 = 4PQ

Q. If $P$ be the sum of odd terms and $Q$ that of even terms in the expansion of $(x + a)^{n}$ , then the value of $[(x + a)^{2 n} - (x - a)^{2 n}]$ equals

$PQ$

$2 PQ$

$4 PQ$

None of these

Q. If P be the sum of odd terms and Q that of even terms in the expansion of (x+a)n, then the value of [(x+a)2n−(x−a)2n] equals

PQ

2PQ

4PQ

None of these

(x+a)n=(t1​+t3​+t5​+...)+(t2​+t4​+t6​+...) ∴(x+a)n=P+Q…(i) and (x−a)n=(t1​+t3​+t5​+...)−(t2​+t4​+t6​+...) (x−a)n=P−Q…(ii) Squaring and subtracting (ii) from (i), we get (x+a)2n−(x−a)2n =(P+Q)2−(P−Q)2=4PQ

Q. If $P$ be the sum of odd terms and $Q$ that of even terms in the expansion of $(x + a)^{n}$ , then the value of $[(x + a)^{2 n} - (x - a)^{2 n}]$ equals

$PQ$

$2 PQ$

$4 PQ$