If S_n=_r=0^n1/^nC_r and t_n=_r=0^nr/^nC_r, then _nS_n is equal to - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

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If S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}} and t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}, then \frac{t _{n}}{S _{n}} is equal to

Solution & Step-by-step Explanation

Given t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}} . Using the property^{n} C_{r} =^{n} C_{n - r}, we can rewrite t_{n} by replacing r with (n - r) : t_{n} = \sum_{r = 0}^{n} \frac{n - r}{^{n} C _{n - r}} = \sum_{r = 0}^{n} \frac{n - r}{^{n} C _{r}} . Adding the two expressions for t_{n} : 2 t_{n} = \sum_{r = 0}^{n} \frac{r + ( n - r )}{^{n} C _{r}} = \sum_{r = 0}^{n} \frac{n}{^{n} C _{r}} . 2 t_{n} = n \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}} 2 t_{n} = n S_{n} . \frac{t _{n}}{S _{n}} = \frac{n}{2} .

Try it yourself before checking the explanation above.

If S_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C _{r}} and t_{n} = \sum_{r = 0}^{n} \frac{r}{^{n} C _{r}}, then \frac{t _{n}}{S _{n}} is equal to

\frac{1}{2} n

\frac{1}{2} n - 1

n - 1

\frac{2 n - 1}{2}

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