Find a formula for $sumlimits_{r=1}^{n} (r^2+1)(r!)$
The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:
- $(n+1)!$
- $(n+2)!-1$
- $ncdot(n+1)!$
- $ncdot(n+2)!$
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
add a comment |
The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:
- $(n+1)!$
- $(n+2)!-1$
- $ncdot(n+1)!$
- $ncdot(n+2)!$
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
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
add a comment |
The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:
- $(n+1)!$
- $(n+2)!-1$
- $ncdot(n+1)!$
- $ncdot(n+2)!$
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
The sum $$sumlimits_{r=1}^{n} (r^2+1)(r!)$$
is equal to:
- $(n+1)!$
- $(n+2)!-1$
- $ncdot(n+1)!$
- $ncdot(n+2)!$
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
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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?
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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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?
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
add a comment |
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?
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
add a comment |
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?
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?
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
add a comment |
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
add a comment |
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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