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Question

Question: \[1 + \frac{4^{2}}{3!} + \frac{4^{4}}{5!} + ......\infty =\]...

1+423!+445!+......∞=1 + \frac{4^{2}}{3!} + \frac{4^{4}}{5!} + ......\infty =

A

e4+e−44\frac{e^{4} + e^{- 4}}{4}

B

e4−e−44\frac{e^{4} - e^{- 4}}{4}

C

e4+e−48\frac{e^{4} + e^{- 4}}{8}

D

e4−e−48\frac{e^{4} - e^{- 4}}{8}

Answer

e4−e−48\frac{e^{4} - e^{- 4}}{8}

Explanation

Solution

log⁡e3\log_{e}3

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} + ......,.