Let s_1 = sum_{j=1}^{10} j(j-1) binom{10}{j}, | vedclass

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

Question

(D) We know that $\sum_{j=0}^{n} \binom{n}{j} = 2^n$.For $s_1 = \sum_{j=2}^{10} j(j-1) \binom{10}{j} = \sum_{j=2}^{10} 10 	imes 9 \binom{8}{j-2} = 90

Vedclass · Accepted Answer

Statement $-1$ is true,Statement $-2$ is false

Vedclass · Answer

(D) We know that $\sum_{j=0}^{n} \binom{n}{j} = 2^n$.For $s_1 = \sum_{j=2}^{10} j(j-1) \binom{10}{j} = \sum_{j=2}^{10} 10 	imes 9 \binom{8}{j-2} = 90 	imes 2^8$.For $s_2 = \sum_{j=1}^{10} j \binom{10}{j} = \sum_{j=1}^{10} 10 \binom{9}{j-1} = 10 	imes 2^9$.For $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j} = \sum_{j=1}^{10} (j(j-1) + j) \binom{10}{j} = s_1 + s_2$.$s_3 = 90 	imes 2^8 + 10 	imes 2^9 = 45 	imes 2^9 + 10 	imes 2^9 = 55 	imes 2^9$.Comparing with the statements:Statement $-1$ is $55 	imes 2^9$,which is true.Statement $-2$ claims $s_2 = 10 	imes 2^8$,but we found $s_2 = 10 	imes 2^9$,so Statement $-2$ is false.Thus,Statement $-1$ is true and Statement $-2$ is false.

Let $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$s_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.
Statement $-1$: $s_3 = 55 \times 2^9$
Statement $-2$: $s_1 = 90 \times 2^8$ and $s_2 = 10 \times 2^8$

Similar Questions

The coefficient of the middle term in the binomial expansion in powers of $x$ of $(1 + \alpha x)^4$ and of $(1 - \alpha x)^6$ is the same if $\alpha$ equals

The least value of $n$ for which the number of integral terms in the Binomial expansion of $(\sqrt[3]{7}+\sqrt[12]{11})^{n}$ is $183$ is:

If $ab \neq 0$ and the sum of the coefficients of $x^7$ and $x^4$ in the expansion of $\left(\frac{x^2}{a}-\frac{b}{x}\right)^{11}$ is $0$,then

The absolute difference of the coefficients of $x^{10}$ and $x^7$ in the expansion of $\left(2x^2+\frac{1}{2x}\right)^{11}$ is equal to

The sum of the rational terms in the binomial expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let $s_1 = \sum_{j=1}^{10} j(j-1) \binom{10}{j}$,$s_2 = \sum_{j=1}^{10} j \binom{10}{j}$,and $s_3 = \sum_{j=1}^{10} j^2 \binom{10}{j}$.Statement $-1$: $s_3 = 55 \times 2^9$Statement $-2$: $s_1 = 90 \times 2^8$ and $s_2 = 10 \times 2^8$