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. 
If you mean $$S_n = (1 + 2 + 3 + \ldots + n)  (3 + 4 + 5 + 6 +7+8+9)$$ the first bracket has a wellknown closed form in terms of $n$ and the second is a constant. 
can that be done? because in a toy with infinite sums. 
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1 + 2 + 10 + 11 + 12 = 36 a = 10, d = 1, n = 5 Formula: 3 + {(n2)*[2*a + d*(n  3)]}/2 Tu comprends? 
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3 + {(52)*[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 
3+infini=10+infini=.... car (10+11+12+....=infini) 
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 
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U1=2 n>=3 Un=7+n Here Sn=(n^2+15n)/214 n>=0 Sn=???? 
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. 
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