$\mathop \sum \limits_{r = 0}^n \frac{{{{\left( { - 1} \right)}^r}}}{{{\;^n}{C_r}}}$ equals

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

$\frac{n}{{n + 1}}\left[ {1 + {{\left( { - 1} \right)}^n}} \right]$

JEE MATHEMATICS MCQ: COMBINATIONS

1 Easy

1 of 1

Application +4 | -1

\sum \limits_{r = 0}^n \frac{{{{\left( { - 1} \right)}^r}}}{{{\;^n}{C_r}}}

equals

View answer and explanation

Correct option: A

\frac{{\left( {n + 1} \right)}}{{\left( {n + 2} \right)}}\left[ {1 + {{\left( { - 1} \right)}^n}} \right]

Option A is correct because the given summation can be simplified using properties of binomial coefficients and telescoping series.

Practice All Chapter MCQs →

m \times n

m

n

A

B

{\;^{n + 2}}{C_8}:{\;^{n - 2}}{P_4} = \frac{{57}}{{16}}

n