Mixing
How do I prove this identity for summations?
$begingroup$
How would I go about proving this identity? I really have no idea where to even start, so any help would be greatly appreciated!
$$sum_{i=0}^n left( i + 1 right) left( left( i + 1 right) ! right) = left( n + 2 right) ! - 1$$
calculus sequences-and-series summation
edited Jan 28 at 3:16
Bladewood
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
$endgroup$
add a comment |
$begingroup$
How would I go about proving this identity? I really have no idea where to even start, so any help would be greatly appreciated!
$$sum_{i=0}^n left( i + 1 right) left( left( i + 1 right) ! right) = left( n + 2 right) ! - 1$$
calculus sequences-and-series summation
edited Jan 28 at 3:16
Bladewood
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
$endgroup$
add a comment |
$begingroup$
How would I go about proving this identity? I really have no idea where to even start, so any help would be greatly appreciated!
$$sum_{i=0}^n left( i + 1 right) left( left( i + 1 right) ! right) = left( n + 2 right) ! - 1$$
calculus sequences-and-series summation
edited Jan 28 at 3:16
Bladewood
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
$endgroup$
How would I go about proving this identity? I really have no idea where to even start, so any help would be greatly appreciated!
$$sum_{i=0}^n left( i + 1 right) left( left( i + 1 right) ! right) = left( n + 2 right) ! - 1$$
calculus sequences-and-series summation
calculus sequences-and-series summation
edited Jan 28 at 3:16
Bladewood
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
edited Jan 28 at 3:16
Bladewood
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
edited Jan 28 at 3:16
Bladewood
335213
edited Jan 28 at 3:16
Bladewood
335213
edited Jan 28 at 3:16
Bladewood
335213
335213
asked Jan 28 at 2:17
etnie1031etnie1031
204
asked Jan 28 at 2:17
etnie1031etnie1031
204
asked Jan 28 at 2:17
etnie1031etnie1031
204
204
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$
Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$
answered Jan 28 at 2:24
abc...abc...
3,237739
$endgroup$
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
add a comment |
$begingroup$
Try using induction by letting $mathrm P(n)$ be the proposition that $$sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$
Your base case $mathrm P(0)$ is $$(1)(1)!=2!-1$$
Now you just have to prove that $mathrm P(k)Rightarrowmathrm P(k+1)$. Can you go from there?
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
$endgroup$
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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$begingroup$
Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$
Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$
answered Jan 28 at 2:24
abc...abc...
3,237739
$endgroup$
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
add a comment |
$begingroup$
Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$
Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$
answered Jan 28 at 2:24
abc...abc...
3,237739
$endgroup$
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
add a comment |
$begingroup$
Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$
Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$
answered Jan 28 at 2:24
abc...abc...
3,237739
$endgroup$
Hint: $(i+1)(i+1)!=(i+2)!-(i+1)!$
Now the sum becomes $$2!-1!+3!-2!+...+(n+1)!-n!+(n+2)!-(n-1)!=(n-2)!-1!=(n-2)!-1$$
answered Jan 28 at 2:24
abc...abc...
3,237739
edited Jan 28 at 2:31
answered Jan 28 at 2:24
abc...abc...
3,237739
answered Jan 28 at 2:24
abc...abc...
3,237739
answered Jan 28 at 2:24
abc...abc...
3,237739
3,237739
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
add a comment |
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
How did you get this? (i+1)(i+1)!=(i+2)!−(i+1)! I thought I am supposed to prove that the sum for n+1 = (n+3)!-1 but I'm having trouble getting there
$endgroup$
– etnie1031
Jan 28 at 3:35
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
Okay I actually got it, using your hint. However, can you explain how you came up with that?
$endgroup$
– etnie1031
Jan 28 at 3:37
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
$begingroup$
@etnie1031 It's common to use telescope sums with especially factorials. Also, you can use induction and sometimes that also works.
$endgroup$
– abc...
Jan 28 at 4:42
add a comment |
$begingroup$
Try using induction by letting $mathrm P(n)$ be the proposition that $$sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$
Your base case $mathrm P(0)$ is $$(1)(1)!=2!-1$$
Now you just have to prove that $mathrm P(k)Rightarrowmathrm P(k+1)$. Can you go from there?
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
$endgroup$
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
add a comment |
$begingroup$
Try using induction by letting $mathrm P(n)$ be the proposition that $$sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$
Your base case $mathrm P(0)$ is $$(1)(1)!=2!-1$$
Now you just have to prove that $mathrm P(k)Rightarrowmathrm P(k+1)$. Can you go from there?
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
$endgroup$
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
add a comment |
$begingroup$
Try using induction by letting $mathrm P(n)$ be the proposition that $$sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$
Your base case $mathrm P(0)$ is $$(1)(1)!=2!-1$$
Now you just have to prove that $mathrm P(k)Rightarrowmathrm P(k+1)$. Can you go from there?
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
$endgroup$
Try using induction by letting $mathrm P(n)$ be the proposition that $$sum_{i=0}^n = (i+1)(i+1)! = (n+2)!$$
Your base case $mathrm P(0)$ is $$(1)(1)!=2!-1$$
Now you just have to prove that $mathrm P(k)Rightarrowmathrm P(k+1)$. Can you go from there?
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
answered Jan 28 at 2:38
Chase Ryan TaylorChase Ryan Taylor
4,45021531
4,45021531
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
add a comment |
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
$begingroup$
Thanks, I get now that I'm supposed to show that P(k+1) = (n+3)!-1. I got to the point where I have ∑i=0n+1 = (n+2)!-1 + (n+2)((n+2)!) . I just can't get that to simplify to (n+3)!-1
$endgroup$
– etnie1031
Jan 28 at 3:32
add a comment |
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