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Informaţii, definiţii, teoreme, formule, exerciţii şi probleme rezolvate din matematica de liceu.

Data publicarii: 03 Martie, 2014

TEORIE

1)\;S_1=1+2+3+\cdots+n=\sum_{k=1}^{k=n}{k}=\frac{n(n+1)}{2}\;\cdot1)\;S_1=1+2+3+\cdots+n=\sum_{k=1}^{k=n}{k}=\frac{n(n+1)}{2}\;\cdot

2)\;S_2=1^2+2^2+3^2+\cdots+n^2=\sum_{k=1}^{k=n}{k^2}=\frac{n(n+1)(2n+1)}{6}\cdot2)\;S_2=1^2+2^2+3^2+\cdots+n^2=\sum_{k=1}^{k=n}{k^2}=\frac{n(n+1)(2n+1)}{6}\cdot

3)\;S_3=1^3+2^3+3^3+\cdots+n^3=\sum_{k=1}^{k=n}{k^3}=\frac{n^2(n+1)^2}{4}\cdot3)\;S_3=1^3+2^3+3^3+\cdots+n^3=\sum_{k=1}^{k=n}{k^3}=\frac{n^2(n+1)^2}{4}\cdot

4)\;\sum_{k=0}^{k=n-1}{[x+\frac{k}{n}]=[nx]},\;\forall{x\in{\mathbb{R}}},\;\forall{n\in{\mathbb{N^{*}}}}\cdot4)\;\sum_{k=0}^{k=n-1}{[x+\frac{k}{n}]=[nx]},\;\forall{x\in{\mathbb{R}}},\;\forall{n\in{\mathbb{N^{*}}}}\cdot   (identitatea lui Hermite)

5)\;(a+b)^n=C_n^0a^nb^0+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+\cdots+C_n^ka^{n-k}b^k+\cdots+C_n^na^0b^n=5)\;(a+b)^n=C_n^0a^nb^0+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+\cdots+C_n^ka^{n-k}b^k+\cdots+C_n^na^0b^n=

=\sum_{k=0}^{k=n}{C_n^ka^{n-k}b^k}\cdot=\sum_{k=0}^{k=n}{C_n^ka^{n-k}b^k}\cdot   (binomul lui Newton)

6)\;\sum_{k=0}^{k=n}{C_n^k}=C_n^0+C_n^1+C_n^2+\cdots+C_n^k+\cdots+C_n^n=2^n\cdot6)\;\sum_{k=0}^{k=n}{C_n^k}=C_n^0+C_n^1+C_n^2+\cdots+C_n^k+\cdots+C_n^n=2^n\cdot

(caz particular al binomului lui Newton, in care a = b = 1) 

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Postat în: SUME-liceu

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