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Question: If \(4\left\lbrack x^{2} + \frac{x^{6}}{3} + \frac{x^{10}}{5} + ..... \right\rbrack = y^{2} + \frac{...

If 4[x2+x63+x105+.....]=y2+y42+y63+......,4\left\lbrack x^{2} + \frac{x^{6}}{3} + \frac{x^{10}}{5} + ..... \right\rbrack = y^{2} + \frac{y^{4}}{2} + \frac{y^{6}}{3} + ......,then.

A

x2y=2x−yx^{2}y = 2x - y

B

x2y=2x+yx^{2}y = 2x + y

C

x=2y2−1x = 2y^{2} - 1

D

x2y=2x+y2x^{2}y = 2x + y^{2}

Answer

x2y=2x−yx^{2}y = 2x - y

Explanation

Solution

Given equation is, 1+22!+223!+234!+.....∞1 + \frac{2}{2!} + \frac{2^{2}}{3!} + \frac{2^{3}}{4!} + .....\infty

1+12!+14!+16!+.....∞1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + .....\infty

12(1+12!+14!+....∞)\frac{1}{2}\left( 1 + \frac{1}{2!} + \frac{1}{4!} + ....\infty \right); 12(1+11!+12!+13!+....∞)\frac{1}{2}\left( 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + ....\infty \right)

On simplification, we get (1+12!+14!+....)(1+13!+15!+....)=\left( 1 + \frac{1}{2!} + \frac{1}{4!} + .... \right)\left( 1 + \frac{1}{3!} + \frac{1}{5!} + .... \right) =.