For which positive integer n do we have
1 + 2 + 3 + 4 + … + n = 1 0 × n ?
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will u plz tell me how it is
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The sum of n consecutive integers (from 1 to n ) is equal to 2 n ( n + 1 ) . Therefore, we have, 2 n ( n + 1 ) = 1 0 n ⇒ n 2 + n = 2 0 n ⇒ n 2 − 1 9 n = 0 Solving gives n = 0 , 1 9 however n is a positive integer so n = 1 9
Please read this note, you will understand:
https://brilliant.org/discussions/thread/discuss-more-about-pyramid-investigations-by-arron/
1 + 2 + 3 + ... + n = 10 × n n(n+1)/2 = 10n (n+1)/2 = 10 n + 1 = 20
We should first know about the identity for the series : 1 + 2 + 3 + . . . . + n = 2 n ( n + 1 ) . So, on applying this in the problem, we get ----
1 + 2 + 3 + . . . . . . + n = 1 0 × n
⟹ 2 n ( n + 1 ) = 1 0 n
⟹ 2 n + 1 = 1 0 ⟹ n + 1 = 2 0 ⟹ n = 1 9
The sum of all integers from 1 to N = ((N+1)/2)*N
If we set that = to 10 * N, then...
(N+1)/2 = 10, and
N+1 = 20, and
N =19
n(n+1)/2=10n
just solve it ... ans 19
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2 n ( n + 1 ) = 1 0 n ⟹ n = 1 9 .