If {}^{20}C_{r} is the coefficient of x^{r} i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.We can wr

Vedclass · Accepted Answer

$420 	imes 2^{18}$

Vedclass · Answer

(D) We know that $\sum_{r=0}^{n} r \cdot {}^{n}C_{r} = n \cdot 2^{n-1}$ and $\sum_{r=0}^{n} r(r-1) \cdot {}^{n}C_{r} = n(n-1) \cdot 2^{n-2}$.We can write $r^{2} = r(r-1) + r$.Therefore,$\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r} = \sum_{r=0}^{20} [r(r-1) + r] \cdot {}^{20}C_{r}$.$= \sum_{r=0}^{20} r(r-1) \cdot {}^{20}C_{r} + \sum_{r=0}^{20} r \cdot {}^{20}C_{r}$.Using the identities with $n=20$:$= 20 	imes 19 	imes 2^{20-2} + 20 	imes 2^{20-1}$.$= 380 	imes 2^{18} + 20 	imes 2^{19}$.$= 380 	imes 2^{18} + 20 	imes 2 	imes 2^{18}$.$= 380 	imes 2^{18} + 40 	imes 2^{18}$.$= (380 + 40) 	imes 2^{18} = 420 	imes 2^{18}$.

If ${}^{20}C_{r}$ is the coefficient of $x^{r}$ in the expansion of $(1+x)^{20}$,then the value of $\sum_{r=0}^{20} r^{2} \cdot {}^{20}C_{r}$ is equal to:

Similar Questions

The value of $\sum_{r=0}^{6} \left({}^{6}C_{r} \cdot {}^{6}C_{6-r}\right)$ is equal to:

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$,then $C_0+2 C_1+3 C_2+\ldots+(n+1) C_n$ is equal to

$C_0 - C_1 + C_2 - C_3 + \dots + (-1)^n C_n$ is equal to

If $\sum_{r=0}^{20} {}^{20+r}C_r = \frac{p}{q} {}^{40}C_{20}$ and $GCD(p, q) = 1$,then $p^2 - q^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module