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Sumele puterilor primelor numere naturale

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Se notează:

S_1(n) = 1+2+ \cdots + n = \sum_{k=1}^n k.
S_2(n) = 1^2+2^2 + \cdots + n^2 = \sum_{k=1}^n k^2.

Având în vedere că (k+1)^2 = k^2 + 2k +1, vom găsi o corelaţie între aceste sume:

\sum_{k=1}^n (k^2+2k+1) = \sum_{k=1}^n (k+1)^2 = \sum_{k=2}^{n+1} k^2 = (n+1)^2 -1 + \sum_{k=1}^n.

Aşadar:

\sum_{k=1}^n k^2 + 2 \sum_{k=1}^n +n= (n+1)^2 -1 \; \Rightarrow \; \sum_{k=1}^n k = \frac {n(n+1)}{2}.

Mergem mai departe pentru calculul S_2(n), unde se va ţine seama că:

k^3 + 3k^2 + 3k +1 = (k+1)^3.

Pentru cazul general, se va ţine seama de relaţia:

(k+1)^p = k^p + C^1_p k^{p-1} + C^2_p k^{p-2} + \cdots + C^p_p k +1 .

Deci:

\sum_{k=1}^p = \sum_{k=1}^n k^p + C^1_p \sum_{k=1}^n k_{p-1} + \cdots + C^p_p \sum_{k=1}^n k + n.

Se va deduce relaţia de recurenţă:

C^1_p S_{p-1}(n) + C^2_p S_{p-2} (n) + \cdots + C^1_p S_1(n) = (n+1)^p-n-1.

Se va obţine pe rând:

S_2(n)= \frac {n(n+1)(2n+1)}{6}
S_3(n)=\left [ \frac {n(n+1)}{2} \right ]^2
S_4(n)= \frac {n(n+1)(2n+1)(3n^2+3n+1)}{30}
S_5(n) = \frac {n^2 (n+1)^2 (2n^2+ 2n -1)}{12}
S_6(n)= \frac {n(n+1)(6n^5 + 15 n^4 + 6 n^3 - 6n^2-n+1)}{42}
S_7(n) = \frac {n^2 (n+1)^2 (3n^4 + 6n^3 -n^2-4n +2)}{24}

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