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#1^1 + 2^2 + ... + n^n = ?

Excited Earthworm answered on April 3, 2022 Popularity 8/10 Helpfulness 1/10

answer #1^1 + 2^2 + ... + n^n = ?

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#1^1 + 2^2 + ... + n^n = ?

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#1^1 + 2^2 + ... + n^2 = ?
n = int(input())
res = (n*(n+1)*(2*n+1)) / 6
print(res)

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Popularity 8/10 Helpfulness 1/10 Language python

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Contributed on Nov 12 2022

Excited Earthworm

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1^2+2^2+3^2+...+n^2

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n(n+1)(2n+1)/6

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Contributed on Oct 05 2022

Muscle Baby

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1^2+2^2+3^2+...+n^2

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1^2+2^2+3^2+...+n^2 = n (n+1) (2n+1)
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                            6

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Contributed on Apr 03 2022

Shristi Gautam

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#1^1 + 2^2 + ... + n^n = ?

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