sum_{k=1}^n k(k+1)(k+2) ldots(k+r-1) = | vedclass

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

Question

(A) Let $S_n = \sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1)$.We use the property of falling factorials: $k(k+1) \ldots (k+r-1) = \frac{k(k+1) \ldots (k+r) -

Vedclass · Accepted Answer

$\frac{n(n+1)(n+2) \ldots(n+r)}{r+1}$

Vedclass · Answer

(A) Let $S_n = \sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1)$.We use the property of falling factorials: $k(k+1) \ldots (k+r-1) = \frac{k(k+1) \ldots (k+r) - (k-1)k(k+1) \ldots (k+r-1)}{r+1}$.Let $f(k) = k(k+1) \ldots (k+r-1)$. Then $f(k) = \frac{1}{r+1} [g(k) - g(k-1)]$,where $g(k) = k(k+1) \ldots (k+r)$.Summing from $k=1$ to $n$ gives a telescoping sum:$S_n = \frac{1}{r+1} \sum_{k=1}^n [g(k) - g(k-1)] = \frac{1}{r+1} [g(n) - g(0)]$.Since $g(0) = 0 \cdot 1 \ldots r = 0$,we have $S_n = \frac{g(n)}{r+1} = \frac{n(n+1)(n+2) \ldots(n+r)}{r+1}$.

$\sum_{k=1}^n k(k+1)(k+2) \ldots(k+r-1) =$

Similar Questions

$2.5 + 5.9 + 8.13 + 11.17 + \ldots$ to $10$ terms $=$

Write the first five terms of the sequence whose $n^{th}$ term is $a_{n} = n \frac{n^{2}+5}{4}$.

The sum of the series $1+3+5^2+7+9^2+\ldots$ up to $40$ terms is equal to

If $S_n = 1^3 + 2^3 + \ldots + n^3$ and $T_n = 1 + 2 + \ldots + n$,then

If $|x| < 1$,what is the sum of the series $1 + 2x + 3x^2 + 4x^3 + \dots \infty$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module