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 February 20th, 2019, 02:35 AM #1 Banned Camp   Joined: Feb 2019 From: France Posts: 8 Thanks: 0 Suite 1 2 10 11 12 13..n Hello everyone, I am looking to calculate the sum Sn = 1 + 2 + 10 + 11 + 12 + 13 + 14 .... n According to n And calculate n according to Sn.
 February 20th, 2019, 02:59 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra If you mean $$S_n = (1 + 2 + 3 + \ldots + n) - (3 + 4 + 5 + 6 +7+8+9)$$ the first bracket has a well-known closed form in terms of $n$ and the second is a constant. Thanks from idontknow
 February 20th, 2019, 03:07 AM #3 Banned Camp   Joined: Feb 2019 From: France Posts: 8 Thanks: 0 can that be done? because in a toy with infinite sums.
February 20th, 2019, 05:46 AM   #4
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Quote:
 Originally Posted by Zou37 I am looking to calculate the sum Sn = 1 + 2 + 10 + 11 + 12 + 13 + 14 .... n
Example:
1 + 2 + 10 + 11 + 12 = 36
a = 10, d = 1, n = 5
Formula:
3 + {(n-2)*[2*a + d*(n - 3)]}/2

Tu comprends?

February 20th, 2019, 09:01 AM   #5
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Quote:
 Originally Posted by Denis Example: 1 + 2 + 10 + 11 + 12 = 36 a = 10, d = 1, n = 5 Formula: 3 + {(n-2)*[2*a + d*(n - 3)]}/2 Tu comprends?
No .

February 20th, 2019, 09:14 AM   #6
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Quote:
 Originally Posted by Denis Example: 1 + 2 + 10 + 11 + 12 = 36 a = 10, d = 1, n = 5 Formula: 3 + {(n-2)*[2*a + d*(n - 3)]}/2
3 + {(n-2)*[2*a + d*(n - 3)]}/2
3 + {(5-2)*[2*10 + 1*(5 - 3)]}/2 = 36

WHAT do you not understand?

The regular arithmetic series is 10 + 11 + 12 .....
What appears before the regular series is 1 + 2
So solve the regular series, then add the
irregular portion which is 1+2 = 3

If you're not sure of the formula, google "arithmetic series".

C'est bon?

EDIT:
note that if a = d = 1, then you have consecutive natural numbers,
so formula can be condensed to n(n + 1)/2; example:
1 + 2 + 3 + 4 + 5 = 15
5*6 / 2 = 15
Those are known as triangular numbers;
for above example: 5th triangular number = 15

Last edited by Denis; February 20th, 2019 at 09:28 AM.

 February 20th, 2019, 09:25 AM #7 Banned Camp   Joined: Feb 2019 From: France Posts: 8 Thanks: 0 3+infini=10+infini=.... car (10+11+12+....=infini)
 February 20th, 2019, 09:48 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 You wanted to calculate the sum "according to n", where n is an integer greater than 9. This can't be infinite. Sn = 3 + (n - 9)(10 + n)/2 = (n² + n - 84)/2
February 20th, 2019, 10:06 AM   #9
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Quote:
 Originally Posted by skipjack You wanted to calculate the sum "according to n", where n is an integer greater than 9. This can't be infinite. Sn = 3 + (n - 9)(10 + n)/2 = (n² + n - 84)/2
U0=1
U1=2
n>=3 Un=7+n

Here Sn=(n^2+15n)/2-14

n>=0 Sn=????

 February 20th, 2019, 10:12 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 You didn't state originally that n is the number of terms. Why define Un? There is no Un in the question or in your solution. Last edited by skipjack; February 20th, 2019 at 10:15 AM.

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