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BITSAT 2020
Exam
BITSAT
Year
2020
Subject
Mathematics
Chapter
Sequences and Series

BITSAT 2020 · Mathematics · Sequences and Series

The sum of the infinite series 2^ 2 2 ! + 2^ 4 4 ! + 2^ 6 6 ! + is...

The sum of the infinite series 222!+244!+266!+…\frac{2^{2}}{2 !}+\frac{2^{4}}{4 !}+\frac{2^{6}}{6 !}+\ldots is equal to

Options

  1. A\mathrme2+12e\frac{\mathrme^{2}+1}{2 \mathrm{e}}
  2. B\mathrme4+12\mathrme2\frac{\mathrme^{4}+1}{2 \mathrme^{2}}
  3. C(\mathrme2−1)22\mathrme2\frac{\left(\mathrme^{2}-1\right)^{2}}{2 \mathrme^{2}}
  4. D(\mathrme2+1)22\mathrme2\frac{\left(\mathrme^{2}+1\right)^{2}}{2 \mathrme^{2}}

Correct answer

C. (\mathrme2−1)22\mathrme2\frac{\left(\mathrme^{2}-1\right)^{2}}{2 \mathrme^{2}}

Step-by-step solution

We know that ex+e−x2=1+x22!+x44!+x66!+…\frac{e^{x}+e^{-x}}{2}=1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\frac{x^{6}}{6 !}+\ldots keeping x=2\mathrm{x}=2, we get Expression =12[e2+e−22]−1=(e2−1)22e2=\frac{1}{2}\left[\frac{e^{2}+e^{-2}}{2}\right]-1=\frac{\left(e^{2}-1\right)^{2}}{2 e^{2}}
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