If C_r denotes the binomial coefficient { }^{ | vedclass

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

Question

(B) Consider the given expression: $S = \sum_{r=0}^n (3r-1) C_r^2$.We know that $C_r^2 = C_r \cdot C_{n-r}$.So,$S = 3 \sum_{r=0}^n r C_r^2 - \sum_{r=0

Vedclass · Accepted Answer

$\left(\frac{3 n-2}{2}\right){ }^{2 n} C_n$

Vedclass · Answer

(B) Consider the given expression: $S = \sum_{r=0}^n (3r-1) C_r^2$.We know that $C_r^2 = C_r \cdot C_{n-r}$.So,$S = 3 \sum_{r=0}^n r C_r^2 - \sum_{r=0}^n C_r^2$.Using the identity $\sum_{r=0}^n C_r^2 = { }^{2n} C_n$ and $r C_r = n { }^{n-1} C_{r-1}$,we have:$S = 3 \sum_{r=1}^n n { }^{n-1} C_{r-1} C_r - { }^{2n} C_n$.$S = 3n \sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r} - { }^{2n} C_n$.The summation $\sum_{r=1}^n { }^{n-1} C_{r-1} { }^{n} C_{n-r}$ represents the coefficient of $x^{n-1}$ in the expansion of $(1+x)^{n-1}(1+x)^n = (1+x)^{2n-1}$,which is ${ }^{2n-1} C_{n-1}$.Thus,$S = 3n { }^{2n-1} C_{n-1} - { }^{2n} C_n$.Since ${ }^{2n-1} C_{n-1} = \frac{n}{2n} { }^{2n} C_n = \frac{1}{2} { }^{2n} C_n$,we get:$S = 3n \left( \frac{1}{2} { }^{2n} C_n \right) - { }^{2n} C_n = \left( \frac{3n}{2} - 1 \right) { }^{2n} C_n = \left( \frac{3n-2}{2} \right) { }^{2n} C_n$.

If $C_r$ denotes the binomial coefficient ${ }^{n} C_r$,then $(-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2$ is equal to

Similar Questions

The coefficient of $t^{20}$ in the expansion of $(1 + t^2)^{10}(1 + t^{10})(1 + t^{20})$ is

The coefficient of $x^{10}$ in the expansion of $(1 + x)^2 (1 + x^2)^3 (1 + x^3)^4$ is equal to

The coefficient of $x^{37}$ in the expansion of $(1-x)^{30} (1 + x + x^2)^{29}$ is:

For the natural numbers $m, n$,if $(1-y)^{m}(1+y)^{n}=1+a_{1} y+a_{2} y^{2}+\ldots +a_{m+n} y^{m+n}$ and $a_{1}=a_{2}=10$,then the value of $(m+n)$ is equal to:

If the coefficient of $x^r$ in the expansion of $(1+x+x^2+x^3)^{100}$ is $a_r$,and $S = \sum_{r=0}^{300} a_r$,then $\sum_{r=0}^{300} r \cdot a_r =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module