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What is the sum of the following sequence:
1.2.3.4.5......+ 2.3.4.5.6......+ 3.4.5.6.7.....4.5.6.7.8.....+996.997.998.999.1000.
Any help would be greatly appreciated. I thank you.

 Jul 9, 2019

Best Answer 

 #10
avatar+23171 
+2

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  \(\dbinom nk\)

 

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

 

laugh

 Jul 11, 2019
 #1
avatar+103999 
+1

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)

.

.

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

 

 

cool cool cool

 Jul 9, 2019
 #2
avatar
0

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

 Jul 9, 2019
 #3
avatar+103999 
0

Thanks, Guest....I was unsure as to what the "dots" were supposed to represent....!!!

 

 

cool cool cool

CPhill  Jul 9, 2019
 #5
avatar+1750 
+1

I know why the guest knew what they meant.  The guest is Mr. BB, the same one who posted the question. Mr. BB often uses unconventional math terminology and symbols.   

I have to say Mr. BB, that’s an optimal method for improving your answering speed, not forgetting your accuracy.  You are much less likely to fuck up the answers when you cherry pick and post your own questions. Of course, you’d know what your own chicken scratch (and crap) means.

 Even with your cherry-picking,   you still manage to bugger the answers.  

 https://web2.0calc.com/questions/combinations-and-permutations_5

 

 

 

GA

GingerAle  Jul 9, 2019
edited by GingerAle  Jul 9, 2019
 #4
avatar
0

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

 Jul 9, 2019
 #6
avatar+103999 
0

Because we can't use the "dots answered" mark......

 

 

cool cool cool 

CPhill  Jul 9, 2019
 #7
avatar+23171 
+1

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

 

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

 

laugh

 Jul 10, 2019
 #8
avatar
0

Use the hockey stick identity to solve this question

 Jul 10, 2019
edited by Guest  Jul 10, 2019
edited by Guest  Jul 10, 2019
 #9
avatar+23171 
+2

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:

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

 

laugh

 Jul 10, 2019
 #10
avatar+23171 
+2
Best Answer

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  \(\dbinom nk\)

 

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

 

laugh

heureka Jul 11, 2019
 #11
avatar+103999 
+1

Thanks, heureka.....I wasn't familiar with the "hockey stick" identity.......very informative....!!!!

 

 

 

cool cool cool

CPhill  Jul 11, 2019
 #12
avatar+23171 
+2

Thank you, CPhill !

 

laugh

heureka  Jul 11, 2019
 #13
avatar
0

Yeah you're welcome

Guest Jul 11, 2019

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