If sum_{k=1}^{30} k left({ }^{30} C _kright)^ | vedclass

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

Question

(C) Let $S = \sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2$. Using the property ${ }^{n} C _k = { }^{n} C _{n-k}$,we have ${ }^{30} C _k = { }^{30} C

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let $S = \sum_{k=1}^{30} k \left({ }^{30} C _kight)^2$. Using the property ${ }^{n} C _k = { }^{n} C _{n-k}$,we have ${ }^{30} C _k = { }^{30} C _{30-k}$. Thus,$S = \sum_{k=1}^{30} k \left({ }^{30} C _{30-k}ight)^2$. Let $j = 30-k$,then $k = 30-j$. As $k$ goes from $1$ to $30$,$j$ goes from $29$ to $0$. $S = \sum_{j=0}^{29} (30-j) \left({ }^{30} C _jight)^2 = 30 \sum_{j=0}^{29} \left({ }^{30} C _jight)^2 - \sum_{j=0}^{29} j \left({ }^{30} C _jight)^2$. Since $30 \left({ }^{30} C _{30}ight)^2 = 30$,we can write $S = 30 \sum_{j=0}^{30} \left({ }^{30} C _jight)^2 - S - 30 \left({ }^{30} C _{30}ight)^2 + 30 \left({ }^{30} C _{30}ight)^2$ is not needed,simply: $2S = 30 \sum_{k=0}^{30} \left({ }^{30} C _kight)^2 - 30 \left({ }^{30} C _{30}ight)^2 + 30 \left({ }^{30} C _{30}ight)^2$ is incorrect. Correct approach: $S = \sum_{k=1}^{30} k \left({ }^{30} C _kight) \left({ }^{30} C _kight) = \sum_{k=1}^{30} k \left({ }^{30} C _kight) \left({ }^{30} C _{30-k}ight) = 30 \sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k}$. Using Vandermonde's Identity,$\sum_{k=1}^{30} { }^{29} C _{k-1} { }^{30} C _{30-k} = { }^{59} C _{29}$. $S = 30 	imes { }^{59} C _{29} = 30 	imes \frac{59!}{29! 30!} = 30 	imes \frac{59! 	imes 30}{30! 30!} = 15 	imes \frac{60!}{30! 30!}$. Thus,$\alpha = 15$.

If $\sum_{k=1}^{30} k \left({ }^{30} C _k\right)^2 = \frac{\alpha 60 !}{(30 !)^2}$,then $\alpha$ is equal to

Similar Questions

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

If $\frac{{}^{11}C_1}{2} + \frac{{}^{11}C_2}{3} + \dots + \frac{{}^{11}C_9}{10} = \frac{n}{m}$ with $\gcd(n, m) = 1$,then $n + m$ is equal to

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

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

If the coefficients of $a^{r-1}$,$a^{r}$,and $a^{r+1}$ in the expansion of $(1+a)^{n}$ are in arithmetic progression,prove that $n^{2}-n(4r+1)+4r^{2}-2=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module