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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) 

7)\;C_n^0+C_n^2+C_n^4+\cdots=C_n^1+C_n^3+C_n^5+\cdots=2^{n-1}\cdot7)\;C_n^0+C_n^2+C_n^4+\cdots=C_n^1+C_n^3+C_n^5+\cdots=2^{n-1}\cdot   

8)\;\sum_{k=0}^{k=n}{(-1)^k}\cdot{C_n^k}=C_n^0-C_n^1+C_n^2-C_n^3+\cdots+(-1)^n\cdot{C_n^n}=0\cdot8)\;\sum_{k=0}^{k=n}{(-1)^k}\cdot{C_n^k}=C_n^0-C_n^1+C_n^2-C_n^3+\cdots+(-1)^n\cdot{C_n^n}=0\cdot  

9)\;a_1+a_2+a_3+\cdots+a_n=a_1+(a_1+r)+(a_1+2r)+\cdots+(a_1+(n-1)r)=9)\;a_1+a_2+a_3+\cdots+a_n=a_1+(a_1+r)+(a_1+2r)+\cdots+(a_1+(n-1)r)=

=\sum_{k=1}^{k=n}{(a_1+(k-1)r)}=\frac{(a_1+a_n)n}{2}\cdot=\sum_{k=1}^{k=n}{(a_1+(k-1)r)}=\frac{(a_1+a_n)n}{2}\cdot  (la progresia aritmetica)

10)\;b_1+b_2+b_3+\cdots+b_n=b_1+b_1q+b_1q^2+\cdots+b_1q^{n-1}=10)\;b_1+b_2+b_3+\cdots+b_n=b_1+b_1q+b_1q^2+\cdots+b_1q^{n-1}=

=\sum_{k=1}^{k=n}{b_1q^{k-1}}={b_1}\cdot{\frac{1-q^n}{1-q}},\;{q}\not=1\cdot=\sum_{k=1}^{k=n}{b_1q^{k-1}}={b_1}\cdot{\frac{1-q^n}{1-q}},\;{q}\not=1\cdot (la progresia geometrica)

11)\;\sum_{k=1}^{k=2n}{\frac{(-1)^{k+1}}{k}}=\sum_{1}^{n}{\frac{1}{n+k}}\cdot11)\;\sum_{k=1}^{k=2n}{\frac{(-1)^{k+1}}{k}}=\sum_{1}^{n}{\frac{1}{n+k}}\cdot   (identitatea lui Catalan)

12)\;\sum_{k=1}^{k=n}{sinkx}=\frac{{sin{\frac{nx}{2}}}\cdot{sin{\frac{(n+1)x}{2}}}}{sin{\frac{x}{2}}},\;x\in{R}\setminus{\{2k\pi|k\in{Z}\}}\cdot12)\;\sum_{k=1}^{k=n}{sinkx}=\frac{{sin{\frac{nx}{2}}}\cdot{sin{\frac{(n+1)x}{2}}}}{sin{\frac{x}{2}}},\;x\in{R}\setminus{\{2k\pi|k\in{Z}\}}\cdot

13)\;\sum_{k=1}^{k=n}{coskx}=\frac{{sin{\frac{nx}{2}}}\cdot{cos{\frac{(n+1)x}{2}}}}{sin{\frac{x}{2}}},\;x\in{R}\setminus{\{2k\pi|k\in{Z}\}}\cdot13)\;\sum_{k=1}^{k=n}{coskx}=\frac{{sin{\frac{nx}{2}}}\cdot{cos{\frac{(n+1)x}{2}}}}{sin{\frac{x}{2}}},\;x\in{R}\setminus{\{2k\pi|k\in{Z}\}}\cdot

Postat în: SUME-liceu

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