If, ${a_n} = \mathop \sum \limits_{r = 0}^n \frac{1}{{{\;^n}{C_r}}}$, then $\mathop \sum \limits_{r = 0}^n \frac{r}{{{\;^n}{C_r}}}$ equals

JEE MATHEMATICS MCQ: BINOMIAL THEOREM

1 Easy

1 of 1

Application +4 | -1

If,

{a_n} = \sum \limits_{r = 0}^n \frac{1}{{{\;^n}{C_r}}}

, then

\sum \limits_{r = 0}^n \frac{r}{{{\;^n}{C_r}}}

equals

View answer and explanation

Correct option: C

\frac{1}{2}n{a_n}

Option C (

\frac{1}{2}n{a_n}

) is correct because by using the property of binomial coefficients and summing the series in reverse, we can derive the given expression.

Practice All Chapter MCQs →

\left({ }^{10} C_0\right)^2-\left({ }^{10} C_1\right)^2+\ldots-\left({ }^{10} C_9\right)^2+\left({ }^{10} C_{10}\right)^2

a

d

\left( {n + 1} \right)

a{C_0} - \left( {a + d} \right){C_1} + \left( {a + 2d} \right){C_2} - ... + ...

(1-x)^{201}\left(1+x+x^2\right)^{200}

\left(\frac{1}{x}+2 x\right)^n

{x^5}

\frac{{1 + {x^2}}}{{1 + x}},\left| x \right| < 1,

\left(\begin{array}{c}2 n+1 \\ 0\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 3\end{array}\right)+\left(\begin{array}{c}2 n+1 \\ 6\end{array}\right)+\ldots=170

n