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Binomial Theorem

Question
CBSEENMA11015669
Wired Faculty App

If space straight S subscript straight n space equals space sum from straight r equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight r end fraction space and space straight t subscript straight n space equals space sum from straight r equals space 0 to straight n of space straight r over straight C subscript straight r space then comma straight t subscript straight n over straight S subscript straight n space is
  • 1 half straight n
  • 1 half straight n minus 1

  • n-1

  • n

Solution

A.

1 half straight n straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction straight S subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space left parenthesis because space to the power of straight n straight C subscript straight r space equals straight C presuperscript straight n subscript straight n minus straight r end subscript right parenthesis ns subscript straight n space equals space sum from straight r space equals 0 to straight n of fraction numerator 1 over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space ns subscript straight n space equals space sum from straight r space equals 0 to straight n of open square brackets fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction plus fraction numerator straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction close square brackets ns subscript straight n space equals space sum from straight r space equals 0 to straight n of space fraction numerator straight n minus straight r over denominator straight C presuperscript straight n subscript straight n minus straight r end subscript end fraction space plus space sum from straight r space equals 0 to straight n of space fraction numerator straight r over denominator straight C presuperscript straight n subscript straight r end fraction ns subscript straight n space equals space straight t subscript straight n plus straight t subscript straight n ns subscript straight n space equals space 2 straight t subscript straight n straight t subscript straight n over straight s subscript straight n space equals straight n over 2

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