Find a formula for $sumlimits_{r=1}^{n} (r^2+1)(r!)$












1














The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:




  1. $(n+1)!$

  2. $(n+2)!-1$

  3. $ncdot(n+1)!$

  4. $ncdot(n+2)!$


My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?






summation







share|cite|improve this question




















  • 5




    Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
    – Winther
    Nov 28 at 9:21












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 28 at 9:23










  • See math.stackexchange.com/questions/576976/…
    – lab bhattacharjee
    Nov 28 at 9:25










  • @RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
    – lab bhattacharjee
    Nov 28 at 10:04
















1














The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:




  1. $(n+1)!$

  2. $(n+2)!-1$

  3. $ncdot(n+1)!$

  4. $ncdot(n+2)!$


My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?






summation







share|cite|improve this question




















  • 5




    Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
    – Winther
    Nov 28 at 9:21












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 28 at 9:23










  • See math.stackexchange.com/questions/576976/…
    – lab bhattacharjee
    Nov 28 at 9:25










  • @RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
    – lab bhattacharjee
    Nov 28 at 10:04














1












1








1


1





The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:




  1. $(n+1)!$

  2. $(n+2)!-1$

  3. $ncdot(n+1)!$

  4. $ncdot(n+2)!$


My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?






summation







share|cite|improve this question















The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:




  1. $(n+1)!$

  2. $(n+2)!-1$

  3. $ncdot(n+1)!$

  4. $ncdot(n+2)!$


My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?






summation




summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 28 at 9:42









Robert Z

92.8k1060131




92.8k1060131










asked Nov 28 at 9:18









Himanshu Shekhar

112




112








  • 5




    Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
    – Winther
    Nov 28 at 9:21












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 28 at 9:23










  • See math.stackexchange.com/questions/576976/…
    – lab bhattacharjee
    Nov 28 at 9:25










  • @RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
    – lab bhattacharjee
    Nov 28 at 10:04














  • 5




    Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
    – Winther
    Nov 28 at 9:21












  • Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    – José Carlos Santos
    Nov 28 at 9:23










  • See math.stackexchange.com/questions/576976/…
    – lab bhattacharjee
    Nov 28 at 9:25










  • @RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
    – lab bhattacharjee
    Nov 28 at 10:04








5




5




Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
– Winther
Nov 28 at 9:21






Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes.
– Winther
Nov 28 at 9:21














Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 at 9:23




Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 at 9:23












See math.stackexchange.com/questions/576976/…
– lab bhattacharjee
Nov 28 at 9:25




See math.stackexchange.com/questions/576976/…
– lab bhattacharjee
Nov 28 at 9:25












@RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
– lab bhattacharjee
Nov 28 at 10:04




@RobertZ, I think people should get a fair idea how to address the current problem from that like math.stackexchange.com/questions/2638073/…
– lab bhattacharjee
Nov 28 at 10:04










1 Answer
1






active

oldest

votes


















4














Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment:
$$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$
then
$$begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\
&=((r+2)! - (r+1)!) - 2((r+1)!-r!).
end{align}$$
Now split the sum in two parts and try to simplify.
What do you obtain?






share|cite|improve this answer























  • does this form an telescoping series?
    – Himanshu Shekhar
    Nov 28 at 9:49










  • $$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
    – lab bhattacharjee
    Nov 28 at 9:51










  • Yes, it is a telescopic sum.
    – Robert Z
    Nov 28 at 10:01













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

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4














Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment:
$$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$
then
$$begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\
&=((r+2)! - (r+1)!) - 2((r+1)!-r!).
end{align}$$
Now split the sum in two parts and try to simplify.
What do you obtain?






share|cite|improve this answer























  • does this form an telescoping series?
    – Himanshu Shekhar
    Nov 28 at 9:49










  • $$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
    – lab bhattacharjee
    Nov 28 at 9:51










  • Yes, it is a telescopic sum.
    – Robert Z
    Nov 28 at 10:01


















4














Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment:
$$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$
then
$$begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\
&=((r+2)! - (r+1)!) - 2((r+1)!-r!).
end{align}$$
Now split the sum in two parts and try to simplify.
What do you obtain?






share|cite|improve this answer























  • does this form an telescoping series?
    – Himanshu Shekhar
    Nov 28 at 9:49










  • $$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
    – lab bhattacharjee
    Nov 28 at 9:51










  • Yes, it is a telescopic sum.
    – Robert Z
    Nov 28 at 10:01
















4












4








4






Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment:
$$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$
then
$$begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\
&=((r+2)! - (r+1)!) - 2((r+1)!-r!).
end{align}$$
Now split the sum in two parts and try to simplify.
What do you obtain?






share|cite|improve this answer














Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment:
$$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$
then
$$begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\
&=((r+2)! - (r+1)!) - 2((r+1)!-r!).
end{align}$$
Now split the sum in two parts and try to simplify.
What do you obtain?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 28 at 10:00

























answered Nov 28 at 9:31









Robert Z

92.8k1060131




92.8k1060131












  • does this form an telescoping series?
    – Himanshu Shekhar
    Nov 28 at 9:49










  • $$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
    – lab bhattacharjee
    Nov 28 at 9:51










  • Yes, it is a telescopic sum.
    – Robert Z
    Nov 28 at 10:01




















  • does this form an telescoping series?
    – Himanshu Shekhar
    Nov 28 at 9:49










  • $$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
    – lab bhattacharjee
    Nov 28 at 9:51










  • Yes, it is a telescopic sum.
    – Robert Z
    Nov 28 at 10:01


















does this form an telescoping series?
– Himanshu Shekhar
Nov 28 at 9:49




does this form an telescoping series?
– Himanshu Shekhar
Nov 28 at 9:49












$$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
– lab bhattacharjee
Nov 28 at 9:51




$$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$
– lab bhattacharjee
Nov 28 at 9:51












Yes, it is a telescopic sum.
– Robert Z
Nov 28 at 10:01






Yes, it is a telescopic sum.
– Robert Z
Nov 28 at 10:01




















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