sum_{r=0}^{10} {}^{40-r} C_5 is equal to | vedclass

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

Question

(B) The given sum is $\sum_{r=0}^{10} {}^{40-r} C_5 = {}^{40} C_5 + {}^{39} C_5 + {}^{38} C_5 + \dots + {}^{30} C_5$. This can be written as $\sum_{k=

Vedclass · Accepted Answer

${}^{41} C_6 - {}^{30} C_6$

Vedclass · Answer

(B) The given sum is $\sum_{r=0}^{10} {}^{40-r} C_5 = {}^{40} C_5 + {}^{39} C_5 + {}^{38} C_5 + \dots + {}^{30} C_5$. This can be written as $\sum_{k=30}^{40} {}^{k} C_5$. Using the identity ${}^{n} C_r + {}^{n} C_{r-1} = {}^{n+1} C_r$,we can rewrite ${}^{k} C_5$ as ${}^{k+1} C_6 - {}^{k} C_6$. Thus,the sum becomes: $\sum_{k=30}^{40} ({}^{k+1} C_6 - {}^{k} C_6) = ({}^{31} C_6 - {}^{30} C_6) + ({}^{32} C_6 - {}^{31} C_6) + \dots + ({}^{41} C_6 - {}^{40} C_6)$. This is a telescoping sum,which simplifies to ${}^{41} C_6 - {}^{30} C_6$.

$\sum_{r=0}^{10} {}^{40-r} C_5$ is equal to

Similar Questions

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $,then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

The value of $\frac{^{100}C_{50}}{51} + \frac{^{100}C_{51}}{52} + \dots + \frac{^{100}C_{100}}{101}$ is:

If $(1 + x)^n = C_0 + C_1x + C_2x^2 + .......... + C_nx^n,$ then $C_0^2 + C_1^2 + C_2^2 + C_3^2 + ...... + C_n^2$ =

Let $Q(x)$ be a polynomial of degree $n$. If $Q(1)=1$ and $\frac{Q(2x)}{Q(x+1)}+\frac{56}{x+7}-8=0$,then the value of ${}^nC_0+{}^nC_1+\ldots+{}^nC_n$ is equal to

$\frac{{^nC_0}}{1} + \frac{{^nC_2}}{3} + \frac{{^nC_4}}{5} + \frac{{^nC_6}}{7} + \dots = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module