Sum of sequence

What is the sum of the following sequence:
\(\mathbf{1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000}\) .

Using the hockey stick identity and the  binomial coefficient  \(\dbinom nk\)

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& \dfrac{5!}{0!} + \dfrac{6!}{1!} + \dfrac{7!}{2!} + \dfrac{8!}{3!} + \dfrac{9!}{4!} + \dfrac{10!}{5!} + \dfrac{11!}{6!} +\ldots + \dfrac{1000!}{995!} \\\\ &=& \dfrac{5!}{5!0!}5! + \dfrac{6!}{5!1!}5! + \dfrac{7!}{5!2!}5! + \dfrac{8!}{5!3!}5! + \dfrac{9!}{5!4!}5! + \dfrac{10!}{5!5!}5! + \dfrac{11!}{5!6!}5! +\ldots + \dfrac{1000!}{5!995!}5! \\\\ &=& \dbinom 55 5! + \dbinom 65 5! + \dbinom 75 5! + \dbinom 85 5! + \dbinom 95 5! + \dbinom {10}5 5! + \dbinom {11}5 5! +\ldots + \dbinom {1000}5 5! \\\\ &=& 5! \left( \underbrace{\dbinom 55 + \dbinom 65 + \dbinom 75 + \dbinom 85 + \dbinom 95 + \dbinom {10}5 + \dbinom {11}5 +\ldots + \dbinom {1000}5}_{\text{long stick of the hockey in Pascals triangle}} \right) \\\\ &=& 5!*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\\\ &=& 5!* \dbinom{1001}{6} \\\\ &=& 5!* \dfrac{1001!}{6!995!} \\\\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}\)

1+ 2 + 3 + 4 + 5  =   15    =   3(5)

2 + 3 + 4 + 5 + 6 =   20   =    4 (5)

3 + 4 +5 + 6 + 7  =   25   =    5(5)

4 + 5 + 6 + 7 + 8  =  30   =    6(5)

.

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996 + 997 + 998 + 999 + 1000 = 998(5)

So...we have

3(5)  + 4(5) + 5(5) + 6(5) + ..... + 998(5)  =

5 ( 3 + 4 + 5 + 6 +  ..... + 998)  =

5  [998 + 3]  [  998 - 3 + 1]  / 2 = 

5 [ 1001] [ 996] / 2  =

5005 * 498  =

2,492,490

CPhill: Why did you sum up the group of "5 integers" instead of multiplying them together as follows:
5!/(5-5)! + 6!/(6-5)! + 7!/(7-5)! + 8!/(8-5)! + 9!(9-5)! + 10!/(10-5)! =5!/0! + 6!/1! + 7!/2! + 8!/3! + 9!/4! + 10!/5! +........+1000!/995!. Which works out as:
∑ [ (n + 4)! / (n - 1)!, n, 1, 996] = 165,170,830,829,004,000

why is the question have the "question answered" mark?

What is the sum of the following sequence:
\(1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000\).

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\ &=& \sum \limits_{k=1}^{996} k(k+1)(k+2)(k+3)(k+4) \\ &=& \sum \limits_{k=1}^{996} k^5 + 10 k^4 + 35 k^3 + 50 k^2 + 24 k \\ &=& \sum \limits_{k=1}^{996} k^5 + 10\sum \limits_{k=1}^{996} k^4 + 35 \sum \limits_{k=1}^{996}k^3 + 50\sum \limits_{k=1}^{996} k^2 + 24\sum \limits_{k=1}^{996} k \\ && \boxed{ \sum \limits_{k=1}^{n} k^1= \dfrac12 n^2 + \dfrac12 n^1 \\ \sum \limits_{k=1}^{n} k^2= \dfrac13 n^3 + \dfrac12 n^2 + \dfrac16 n^1 \\ \sum \limits_{k=1}^{n} k^3= \dfrac14 n^4 + \dfrac12 n^3 + \dfrac14 n^2 \\ \sum \limits_{k=1}^{n} k^4= \dfrac15 n^5 + \dfrac12 n^4 + \dfrac13 n^3 -\dfrac1{30} n^1 \\ \sum \limits_{k=1}^{n} k^5= \dfrac16 n^6 + \dfrac12 n^5 + \dfrac{5}{12} n^4 -\dfrac1{12} n^2 } \\ &=& \dfrac16 996^6 + \dfrac12 996^5 + \dfrac{5}{12} 996^4 -\dfrac1{12} 996^2 + 10\left(\dfrac15 996^5 + \dfrac12 996^4 + \dfrac13 996^3 -\dfrac1{30} 996 \right) \\ &&+ 35\left( \dfrac14 996^4 + \dfrac12 996^3 + \dfrac14 996^2 \right) + 50\left( \dfrac13 996^3 + \dfrac12 996^2 + \dfrac16 996 \right) \\ &&+ 24\left( \dfrac12 996^2 + \dfrac12 996 \right) \\ &=& \dfrac16 996^6 \\ && + \dfrac12 996^5 + \dfrac{10}5 996^5 \\ && + \dfrac{50}{3} 996^4+ \dfrac{10}2 996^4+ \dfrac{35}4 996^4 \\ && + \dfrac{10}3 996^3+ \dfrac{35}2 996^3+ \dfrac{50}3 996^3 \\ && -\dfrac1{12} 996^2 + \dfrac{35}4 996^2 + \dfrac{50}2 996^2 + \dfrac{24}2 996^2 \\ && -\dfrac{10}{30} 996 + \dfrac{50}6 996 + \dfrac{24}2 996 \\ &=& \dfrac16 996^6 + \dfrac{5}{2}\cdot 996^5 + \dfrac{85}{6} 996^4 + \dfrac{75}2 996^3 + \dfrac{274}6 996^2 +20\cdot 996 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array}\)

Use the hockey stick identity to solve this question

What is the sum of the following sequence:
\(1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000\) .

Using the hockey stick identity:

\(\begin{array}{|rcll|} \hline &&\mathbf{ 1\cdot 2\cdot 3\cdot 4\cdot 5 + 2\cdot 3\cdot 4\cdot 5\cdot 6 + 3\cdot 4\cdot 5\cdot 6\cdot 7 + 4\cdot 5\cdot 6\cdot 7\cdot 8 +\ldots + 996\cdot 997\cdot 998\cdot 999\cdot 1000 } \\\\ &=& 120 + 720 + 2510 + 6720 + 15120 + 30240 + 55440 +\ldots + 990034950024000 \\ &=& 120 +6*120 + 21*120 + 56*120 + 126*120 + 252*120 + 462*120 +\ldots + 8250291250200*120 \\ &=& 120*(\underbrace{1+ 6 + 21 + 56 + 126 + 252 + 462 +\ldots + 8250291250200}_{\text{long stick of the hockey in Pascals triangle}} ) \\ &=& 120*\left(\dbinom{5}{5}+\dbinom{6}{5}+\dbinom{7}{5}+\dbinom{8}{5}+\dbinom{9}{5}+\dbinom{10}{5}+\dbinom{11}{5}+\ldots +\dbinom{1000}{5} \right) \\ &=& 120*\left(\underbrace{\dbinom{1001}{6}}_{\text{bend of the hockey stick(hockey stick identity)} } \right) \\ &=& 120* \dbinom{1001}{6} \\ &=& 120*1376423590241700 \\ &=& \mathbf{165170830829004000} \\ \hline \end{array} \)