Question

Calcule a soma dos n primeiros termos da PA
(1-n/n, 2-n/n, 3-n/n,...)

Answer 1

Seja $S$ a soma procurada

$S=\dfrac{1-n}{n}+\dfrac{2-n}{n}+...+\dfrac{n-n}{n}=\\\\\dfrac{1}{n} \cdot [(1-n)+(2-n)+...+(n-n)]=\\\\\dfrac{1}{n} \cdot [(1+2+3+...+n)-n^2]$

Sabemos que pelas somas dos termos da PA que

$1+2+3+...+n=\dfrac{n(1+n)}{2}$

Portanto

$S=\dfrac{1}{n} \left [\dfrac{n(n+1)}{2}-n^2\right ]\Rightarrow \boxed {S=\dfrac{1-n}{2}}$