IF S n - r -0 n 1- CC r and t n - r -0 n r - n r- then t n - S n is equal toA- n -1B- 2 n -1-2C- 1-2 n-1D- 1-2 n

The correct option is D 1/2nS_n= 1/^nC_0+1/^nC_1+1/^nC_2+ ⋯ + 1/^nC_n t_n = 0/^nC_0+ 1/^nC_1+2/^nC_2+⋯ + n/^nC_n t_n= n/^nC_n+ n 1/^nC_n 1+n 2/^nC_n 2+ ⋯ + 0 ...

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Byju's Answer

Standard XI

Mathematics

Summation by Sigma Method

IF S n =∑ r =...

Question

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{2 n - 1}{2}

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\frac{1}{2} n - 1

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n - 1

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\frac{1}{2} n

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Solution

The correct option is D $\frac{1}{2} n$
$S_{n} = \frac{1}{^{n} C_{0}} + \frac{1}{^{n} C_{1}} + \frac{1}{^{n} C_{2}} + \dots + \frac{1}{^{n} C_{n}} t_{n} = \frac{0}{^{n} C_{0}} + \frac{1}{^{n} C_{1}} + \frac{2}{^{n} C_{2}} + \dots + \frac{n}{^{n} C_{n}} t_{n} = \frac{n}{^{n} C_{n}} + \frac{n - 1}{^{n} C_{n - 1}} + \frac{n - 2}{^{n} C_{n - 2}} + \dots + \frac{0}{^{n} C_{0}}$
Add, $2 t_{n} = (n) [\frac{1}{^{n} C_{0}} + \frac{1}{^{n} C_{1}} + \dots \frac{1}{^{n} C_{n}}] = n S_{n} ∴ \frac{t_{n}}{S_{n}} = \frac{n}{2}$

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IF Sn=∑nr=01nCr and tn=∑nr=0rnCr, then tnSn is equal to

The correct option is D 12nSn=1nC0+1nC1+1nC2+⋯+1nCntn=0nC0+1nC1+2nC2+⋯+nnCntn=nnCn+n−1nCn−1+n−2nCn−2+⋯+0nC0 Add, 2tn=(n)[1nC0+1nC1+⋯1nCn]=nSn∴tnSn=n2

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