Q49 / 110 · VITEEE 2023 · Mathematics

The value of \(\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots \ldots\). is equal to

  • A
    \(\frac{2^{\mathrm{n}}+1}{\mathrm{n}+1}\)
  • B
    \(\frac{2^{\mathrm{n}}}{\mathrm{n}+1}\)
  • C
    \(\frac{2^{\mathrm{n}}+1}{\mathrm{n}-1}\)
  • Correct answer
    D
    \(\frac{2^{\mathrm{n}}-1}{\mathrm{n}+1}\)

Solution

\(\begin{aligned} & \frac{\mathrm{C}_1}{2}+\frac{\mathrm{C}_3}{4}+\frac{\mathrm{C}_5}{6}+\ldots . \\ & =\frac{\mathrm{n}}{2}+\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{4 !} \\ & +\frac{\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)(\mathrm{n}-3)(\mathrm{n}-4)}{6 !}+\ldots \ldots \\ & =\frac{1}{\mathrm{n}+1}\left[\frac{(\mathrm{n}+1) \mathrm{n}}{2 !}+\frac{(\mathrm{n}+1) \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)}{4 !}+\right. \\ & \left.\frac{(n+1) n(n-1)(n-2)(n-3)(n-4)}{6 !}+\ldots .\right] \\ & =\frac{1}{\mathrm{n}+1}\left[{ }^{\mathrm{n}+1} \mathrm{C}_2+{ }^{\mathrm{n}+1} \mathrm{C}_4+{ }^{\mathrm{n}+1} \mathrm{C}_6+\ldots . .\right] \\ & =\frac{1}{\mathrm{n}+1}\left[2^{\mathrm{n}+1-1}-{ }^{\mathrm{n}+1} \mathrm{C}_0\right]=\frac{2^{\mathrm{n}}-1}{\mathrm{n}+1} \\ & \end{aligned}\)

Hence, the correct answer is option: 'D'.

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