\({1\over2!}+{1\over4!}+{1\over6!}+\ ...\ \infty\) का मान क्या है?

  1. e2
  2. \(\rm{e\ -\ e^{-1}\over2}\)
  3. \(\rm{e\ +\ e^{-1}\over2}\)
  4. इनमें से कोई नहीं 

Answer (Detailed Solution Below)

Option 4 : इनमें से कोई नहीं 
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Detailed Solution

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संकल्पना:

चरघातांकी श्रृंखला:

  • ex = \(\rm1+{x\over1!}+{x^2\over2!}+{x^3\over3!}+\ ...\ \infty\).

गणना:

हम जानते हैं कि ex = \(\rm1+{x\over1!}+{x^2\over2!}+{x^3\over3!}+\ ...\ \infty\)

साथ ही, e-x = \(\rm1-{x\over1!}+{x^2\over2!}-{x^3\over3!}+\ ...\ \infty\)

∴ ex + e-x\(\rm2+{2x^2\over2!}+{2x^4\over4!}+{2x^6\over6!}+\ ...\ \infty\)

x = 1 रखने पर, हमें निम्न प्राप्त होता है:

\(\rm1+{1\over2!}+{1\over4!}+{1\over6!}+\ ...\ \infty={e\ +\ e^{-1}\over2}\)

⇒ \(\rm{1\over2!}+{1\over4!}+{1\over6!}+\ ...\ \infty={e\ +\ e^{-1}\over2}-1\)

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