The sum of series $\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots$ is
The sum of series $\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots$ is
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$\frac{\left(\mathrm{e}^2-1\right)}{2}$
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$\frac{(e-1)^2}{2 e}$
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$\frac{\left(\mathrm{e}^2-1\right)}{2 \mathrm{e}}$
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$\frac{\left(e^2-2\right)}{e}$
Solution
$\frac{\mathrm{e}^\alpha+\mathrm{e}^{-\alpha}}{2}=1+\frac{\alpha^2}{2 !}+\frac{\alpha^4}{4 !}+\frac{\alpha^6}{6 !}+\ldots \ldots$
$\frac{\mathrm{e}^\alpha+\mathrm{e}^{-\alpha}}{2}-1=\frac{\alpha^2}{2 !}+\frac{\alpha^4}{4 !}+\frac{\alpha^6}{6 !}+\ldots \ldots$
put $\alpha=1$, we get
$\frac{(\mathrm{e}-1)^2}{2 \mathrm{e}}=\frac{1}{2 !}+\frac{1}{4 !}+\frac{1}{6 !}+\ldots \ldots \ldots \ldots$
Asked in: JEE Main 2004
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