Question

Considere uma progressão aritmética, cuja lei do termo geral é

$\mathsf{a_n}$ = a₁ + (n – 1) · r,

com n natural, n ≥ 1.

Usando a lei telescópica para os somatórios
$\mathsf{\displaystyle\sum_{k=p}^q}$ [f(k + 1) – f(k)] = f(q + 1) – f(p), com p ≤ q,

deduza a fórmula para a soma dos n primeiros termos da P.A.:
$\mathsf{S_n=\displaystyle\sum_{k=1}^n a_k=\frac{(a_1+a_n)\cdot n}{2}}.$

Answer 1

Definindo $\Delta(a_{k})=a_{k+1}-a_{k}$ para uma sequência $a_{k}$ qualquer, temos que

$\Delta(k)=(k+1)-k=1~~\Rightarrow~~\displaystyle\sum_{k=1}^{n}1=\sum_{k=1}^{n}\Delta(k)=(n+1)-1=n$
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$\displaystyle\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}\big[a_{1}+(k-1)\cdot r\big]=\sum_{k=1}^{n}a_{1}+\sum_{k=1}^{n}(k-1)\cdot r\\\\\\=a_{1}\sum_{k=1}^{n}1+\sum_{k=1}^{n}(kr-r)=a_{1}\cdot n+\sum_{k=1}^{n}kr-\sum_{k=1}^{n}r\\\\\\=na_{1}-r\sum_{k=1}^{n}1+\sum_{k=1}^{n}kr=na_{1}-r\cdot n-r\sum_{k=1}^{n}k\\\\\\=n(a_{1}-r)-r\sum_{k=1}^{n}k$

Já vimos que diferenças de polinômios de grau $n$ são polinômios de grau $n-1$, então, vamos encontrar $\Delta\big(k^{2}\big)$:

$\Delta(k^{2})=(k+1)^{2}-k^{2}=(k+1+k)\cdot(k+1-k)=2k+1$

Isolando $k$ na expressão acima:

$\boxed{k=\frac{1}{2}\big[\Delta(k^{2})-1\big]}$

Retomando, temos

$\displaystyle\sum_{k=1}^{n}a_{k}=n(a_{1}-r)+r\sum_{k=1}^{n}k\\\\\\\sum_{k=1}^{n}a_{k}=n(a_{1}-r)+r\sum_{k=1}^{n}\frac{1}{2}\big[\Delta(k^{2})-1\big]\\\\\\\sum_{k=1}^{n}a_{k}=n(a_{1}-r)+\frac{r}{2}\bigg[\sum_{k=1}^{n}\Delta(k^{2})-\sum_{k=1}^{n}1\bigg]\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{2n(a_{1}-r)}{2}+\dfrac{r}{2}\bigg[\sum_{k=1}^{n}\Delta(k^{2})-n\bigg]$

Pela propriedade telescópica, temos que

$\displaystyle\sum_{k=1}^{n}\Delta(a_{k})=a_{n+1}-a_{1}=(n+1)^{2}-1^{2},\,\,\,\mathsf{onde~a_{k}=k^{2}}$

Então:

$\displaystyle\sum_{k=1}^{n}a_{k}=\dfrac{2n(a_{1}-r)}{2}+\dfrac{r}{2}\bigg[(n+1)^{2}-1^{2}-n\bigg]\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{2n(a_{1}-r)}{2}+\dfrac{r}{2}\bigg[n^{2}+2n+1-1-n\bigg]\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{2n(a_{1}-r)}{2}+\dfrac{r(n^{2}+n)}{2}\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{2n(a_{1}-r)+r(n^{2}+n)}{2}$

Colocando $n$ em evidência no numerador:

$\displaystyle\sum_{k=1}^{n}a_{k}=\dfrac{n\cdot\big[2(a_{1}-r)+r(n+1)\big]}{2}\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{(2a_{1}-2r+nr+r)\cdot n}{2}\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{(a_{1}+nr-r+a_{1})\cdot n}{2}\\\\\\\sum_{k=1}^{n}a_{k}=\dfrac{(a_{1}+(n-1)\cdot r+a_{1})\cdot n}{2}$

Como $a_{1}+(n-1)\cdot r=a_{n}$, concluímos que

$\boxed{\boxed{\sum_{k=1}^{n}a_{k}=\dfrac{(a_{1}+a_{n})\cdot n}{2}}}$