Question

Determinați numărul de termeni ai sumei, știind că 1+3+5+ ...+n= 576.

rapid va rog

Answer 1

 

$\displaystyle\bf\\1+3+5+ ...+n= 576\\\\Se~cere~numarul~de~termeni~ai~sumei.\\\\Metoda~1~~(metoda~usoara):\\\\Suma~dumerelor~impare~care~incep~de~la~1~este~un~generator\\de~patrate~perfecte.\\Exemple:\\1=1^2=pp\\1+3=4=2^2=pp\\1+3+5=9=3^2=pp\\s.a.m.d.\\\\Suma~acestor~numere~este~T^2\\ unde~T=numarul~de~termeni~din~sir.\\\\1+3+5+ ...+n= 576\\\\576=24^2=pp\\\\\implies~T=24~de~termeni.$

.

$\displaystyle\bf\\Metoda~2~~(metoda~grea):$

.

$\displaystyle\bf\\Calculam~numarul~de~termeni~(T)~in~functie~de~n.\\\\T=\frac{n-1}{2}+1=\frac{n-1+2}{2}=\frac{n+1}{2}~termeni\\\\Calculam~suma~cu~Gauss:\\\\ S=1+3+5+ ...+n= 576=\frac{T(n+1)}{2}=\frac{(\frac{n+1}{2})(n+1)}{2}=\\\\=\frac{(n+1)(n+1)}{4}=\frac{(n+1)^2}{4}=\frac{(n+1)^2}{2^2}=\left(\frac{(n+1)}{2}\right)^2=576\\\\\left(\frac{(n+1)}{2}\right)^2=24^2\\\\\implies~\frac{(n+1)}{2}=24\\\\n+1=48\\n=47\\\\T=\frac{n+1}{2}=\frac{47+1}{2}=\frac{48}{2}=24~de~termeni$