JEE Main 2022 · Mathematics · Binomial Theorem · medium

The value of (1)/(2^8Σr=1^10 r^2binom10r is

Numerical answer

The value of

\dfrac{1}{2^{8}}\displaystyle\sum_{r=1}^{10} r^2\binom{10}{r}

Show answer and solution

Answer: 110

Write

r^2=r(r-1)+r

. Using

r\binom{n}{r}=n\binom{n-1}{r-1}

and

r(r-1)\binom nr=n(n-1)\binom{n-2}{r-2}

we get

\sum r(r-1)\binom{10}{r}=10\cdot9\cdot2^{8}

and

\sum r\binom{10}{r}=10\cdot2^{9}

. Therefore

\sum r^2\binom{10}{r}=90\cdot2^8+10\cdot2^9=2^8(90+20)=110\cdot2^8

. Dividing by

2^8

gives

110

. This agrees with the general formula

n(n+1)2^{n-2}

for

n=10

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake