Efectueaza o cautare in website!

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

TEORIE

Data publicarii: 03.03.2014

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) 

CONTINUARE LA : TEORIE

EXERCITIUL 13

Data publicarii: 06.11.2014

Suport teoretic:

Inegalitati,sume,integrale definite,primitive directe,progresii,ecuatii trigonometrice.

Enunt:

Sa se rezolve ecuatia:

\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\frac{1-sin^nx}{2sinx},\;unde\;{x}\in{(0,\frac{\pi}{2})},\;{t}>{0},\;{n}\in{\mathbb{N^{*}}}.\sum_{k=0}^{k=n-1}{\big(\int_0^{sinx}{(k+1)t^k}dt\big)}=\frac{1-sin^nx}{2sinx},\;unde\;{x}\in{(0,\frac{\pi}{2})},\;{t}>{0},\;{n}\in{\mathbb{N^{*}}}.

Raspuns:

x = π/6.

CONTINUARE LA : EXERCITIUL 13

EXERCITIUL 12

Data publicarii: 06.11.2014

Suport teoretic:

Limite de siruri,sume,logaritmi,Weierstrass,functii derivabile,Stolz-Césaro,l'Hospital.

Enunt:

Se definesc sirurile (an) si (bn), unde nЄN*\{1}, astfel incat:

a_n=\sum_{i=2}^{i=n}{(\sum_{j=2}^{j=i}{ln(\frac{j}{j-1})})}\;si\;b_n=\frac{a_n}{ln(n+1)!}.a_n=\sum_{i=2}^{i=n}{(\sum_{j=2}^{j=i}{ln(\frac{j}{j-1})})}\;si\;b_n=\frac{a_n}{ln(n+1)!}.

Sa se calculeze L = lim(bn).

Raspuns:

L = 1.

CONTINUARE LA : EXERCITIUL 12

EXERCITIUL 11

Data publicarii: 01.11.2014

Suport teoretic:

Sume trigonometrice,ecuatii trigonometrice,functii trigonometrice inverse,identitati

trigonometrice.

Enunt:

Sa se rezolve in multimea numerelor naturale, mai mari sau egale cu 2, ecuatia:

\sum_{k=1}^{k=x-1}(\arcsin{\frac{1}{k}}+\arccos{\frac{1}{k+1}})=2011\cdot\frac{\pi}{2} +\arcsin{\frac{1}{x}}.\sum_{k=1}^{k=x-1}(\arcsin{\frac{1}{k}}+\arccos{\frac{1}{k+1}})=2011\cdot\frac{\pi}{2} +\arcsin{\frac{1}{x}}.

Raspuns:

x = 2012.

CONTINUARE LA : EXERCITIUL 11

EXERCITIUL 10

Data publicarii: 15.10.2014

Suport teoretic:

Descompuneri in factori,sume de puteri.

Enunt:

Sa se calculeze suma

{S_n} = 5 + 55 + 555 +\cdots+\begin{matrix} \underbrace{555\cdots5 }\\n\end{matrix}{S_n} = 5 + 55 + 555 +\cdots+\begin{matrix} \underbrace{555\cdots5 }\\n\end{matrix}

si apoi sa se verifice rezultatul gasit folosind metoda inductiei matematice.

Raspuns:

S_n=\frac{50(10^n-1)-45n}{81}.S_n=\frac{50(10^n-1)-45n}{81}.

CONTINUARE LA : EXERCITIUL 10

 

CATEGORII :


Arhiva blog-ului

 

 
Developed by Hagau Ioan