Question

Cum stabilesc o formula pentru suma:
S=1*2+2*3+3*4+...+n*(n+1).

Answer 1

Răspuns

Explicație pas cu pas:

Answer 2

 

$\displaystyle\\S=1\times2+2\times3+3\times4+...+n\times(n+1)\\\\n\times(n+1)=n^2+n\\\\S=1\times2+2\times3+3\times4+...+n\times(n+1)\\\\S=(1^2+1)+(2^2+2)+(3^2+3)+...+(n^2+n)\\\\\text{Observam ca avem 2 feluri de termeni care se aduna.}\\\\\text{\^Ii separam in 2 subsiruri pentrucare avem formule.}$

$\displaystyle\\\\S=(1^2+2^2+3^2+...+n^2)+(1+2+3+...+n)\\\\\text{Pentru primul sir avem formula:}\\\\1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\\\\\text{Pentru sirul al 2-lea avem formula lui Gauss:}\\\\1+2+3+...+n=\frac{n(n+1)}{2}\\\\\text{Suma initiala devine:}\\\\S=1\cdot2+2\cdot3+3\cdot4+...+n\cdot(n+1)=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$