If an - -r-0n 1-nCr then -r-0n r-nCr equals

Let b = ∑_r=0^n r/^nC_r = ∑_r=0^n n (n r)/^nC_r = n ∑_r=0^n 1/^nC_r ∑_r=0^n n r/^nC_r=na_n ∑_r=0^n n r/^nC_n r [∵ ^nC_r = ^nC_n r] =na_n b ⇒ 2b =na_n ⇒ ...

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Byju's Answer

Standard XIII

Mathematics

Binomial Coefficients

If an = ∑r=0n...

Question

If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}$ then $\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ equals

(n - 1) a_{n}

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n a_{n}

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\frac{1}{2} n a_{n}

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None of these

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Solution

The correct option is C $\frac{1}{2} n a_{n}$
Let $b = \sum_{r = 0}^{n} \frac{r}{^{n} C_{r}} = \sum_{r = 0}^{n} \frac{n - (n - r)}{^{n} C_{r}}$
$= n \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}} - \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{r}} = n a_{n} - \sum_{r = 0}^{n} \frac{n - r}{^{n} C_{n - r}} [∵^{n} C_{r} =^{n} C_{n - r}] = n a_{n} - b \Rightarrow 2 b = n a_{n} \Rightarrow b = \frac{n}{2} a_{n}$

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If an=∑nr=01nCr then ∑nr=0rnCr equals

The correct option is C 12nanLet b=∑nr=0rnCr=∑nr=0n−(n−r)nCr =n∑nr=01nCr−∑nr=0n−rnCr=nan−∑nr=0n−rnCn−r[∵ nCr=nCn−r]=nan−b⇒ 2b=nan ⇒ b=n2an

If $a_{n} = \sum_{r = 0}^{n} \frac{1}{^{n} C_{r}}$ then $\sum_{r = 0}^{n} \frac{r}{^{n} C_{r}}$ equals