1-2-1-4-1-6-1-1-3-1-5-1-7- is equal to

The correct option is D e 1e+1 1/2!+1/4!+1/6!+...∞1+1/3!+1/5!+1/7!+...∞ =2{1/2!+1/4!+1/6!+...∞}2{ 1+1/3!+1/5!+...∞ #160;}=(e+e^ 1) 2(e e^ 1) ...

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Byju's Answer

Standard XII

Mathematics

Sequence

1/2!+1/4!+1/6...

Question

$\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + . . . \infty}{1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + . . . \infty}$ is equal to

\frac{e + 1}{e - 1}

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\frac{e - 1}{e + 1}

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\frac{e^{2} + 1}{e^{2} - 1}

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\frac{e^{2} - 1}{e^{2} + 1}

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Solution

The correct option is D $\frac{e - 1}{e + 1}$
$\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + . . . \infty}{1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + . . . \infty}$

$= \frac{2 {\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + . . . \infty}}{2 {1 + \frac{1}{3!} + \frac{1}{5!} + . . . \infty}} = \frac{(e + e^{- 1}) - 2}{(e - e^{- 1})}$

$= \frac{e + \frac{1}{e} - 2}{e - \frac{1}{e}}$
$= \frac{e^{2} + 1 - 2 e}{e^{2} - 1}$

$= \frac{(e - 1)^{2}}{(e - 1) (e + 1)}$

$= \frac{e - 1}{e + 1}$

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12!+14!+16!+...∞1+13!+15!+17!+...∞ is equal to

The correct option is D e−1e+112!+14!+16!+...∞1+13!+15!+17!+...∞=2{12!+14!+16!+...∞}2{1+13!+15!+...∞}=(e+e−1)−2(e−e−1)=e+1e−2e−1e=e2+1−2ee2−1=(e−1)2(e−1)(e+1)=e−1e+1

$\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + . . . \infty}{1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + . . . \infty}$ is equal to