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

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(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) -

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

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(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 $S_n = \frac{g(n)}{r+1} = \frac{n(n+1)(n+2) \ldots(n+r)}{r+1}$.

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

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