First slide

Binomial theorem for positive integral Index

Question

In the expansion of (x + a)ⁿ if the sum of odd terms is P and the sum of even terms is Q, then

Moderate

A

$P^{2} - Q^{2} = {(x^{2} - a^{2})}^{n}$
B

$4 PQ = (x + a)^{2 n} - (x - a)^{2 n}$
C

$2 (P^{2} + Q^{2}) = (x + a)^{2 n} + (x - a)^{2 n}$
D

none of these

Solution

We have
$(x + a)^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} a +^{n} C_{2} x^{n - 2} a^{2} + \dots +^{n} C_{n} a^{n}$
$= [^{n} C_{0} x^{n} +^{n} C_{2} x^{n - 2} a^{2} + \dots] + [^{n} C_{1} x^{n - 1} a +^{n} C_{3} x^{n - 3} a^{3} + \dots]$
or $(x + a)^{n} = P + Q$ (1)
Similarly,
$(x - a)^{n} = P - Q$ (2)
(i) $(1) \times (2) \Rightarrow P^{2} - Q^{2} = {(x^{2} - a^{2})}^{n}$
(ii) Squaring (1) and (2) and substracting (2) from (1), we get
$4 PQ = (x + a)^{2 n} - (x - a)^{2 n}$
(iii) Squaring (1) and (2) and adding,
$2 (P^{2} + Q^{2}) = (x + a)^{2 n} + (x - a)^{2 n}$

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