Question

if the sum of odd terms and the sum of even terms in X + a whole power n are p and q respectively then 4 PQ equal to​

Answer 1

Answer:

We know that

$(x + a) {}^{n} = \binom{n}{0} {x}^{0} + \binom{n}{1} {x}^{1} + \binom{n}{2} {x}^{2} + .... \binom{n}{n} {x}^{n} \\ \\ (a - x) {}^{n} = \binom{n}{0} { x}^{0} - \binom{n}{1} {x}^{1} + \binom{n}{2} {x}^{2} + .... \binom{n}{n} {x}^{n}$

First add both equations , you will see all even terms will cancel each out . only odd will remain .

Similarly subtract both and all odd terms will be vanished . So

$(a + x) {}^{n} + (a - x) {}^{n} = 2p \\ \\ (a + x) {}^{n} - (a - x) {}^{n} = 2q$

Multiply both

$\{(a + x) {}^{n} + (a - x) {}^{n} \} \{ (a + x) {}^{n} - (a - x) {}^{n} \} = 4p q \\ \\ \red{ {\boxed{ (a + x) {}^{2n} - (a - x) {}^{2n} = 4p q}}}$