Somas telescópicas

Suponhamos que o somando $a_{i}$ é decomponível numa diferença $A_{i+1}-A_{i}$. Então,

$\displaystyle\sum_{i=0}^{n}a_{i}=\sum_{i=0}^{n}A_{i+1}-A_{i}=\sum_{i=0}^{n}A_{i+1}-\sum_{i=0}^{n}A_{i}$.

Como

$\displaystyle\sum_{i=0}^{n}A_{i+1}=\sum_{i=1}^{n+1}A_{i}$$=\left(\displaystyle\sum_{i=1}^{n}A_{i}\right) +A_{n+1}=A_{n+1}+\displaystyle\sum_{i=1}^{n}A_{i}$,

e

$\displaystyle\sum_{i=0}^{n}A_{i}=A_{0}+\sum_{i=0}^{n}A_{i}$

vem

$\displaystyle\sum_{i=0}^{n}a_{i}=A_{n+1}+\sum_{i=1}^{n}A_{i}-A_{0}-\sum_{i=0}^{n}A_{i}=A_{n+1}-A_{0}$.

Analogamente,

$\displaystyle\sum_{i=m}^{n}a_{i} =\sum_{i=m}^{n}A_{i+1}-A_{i}=\sum_{i=m}^{n}A_{i+1}-\sum_{i=m}^{n}A_{i}$$\displaystyle=\sum_{i=m+1}^{n+1}A_{i}-\sum_{i=m}^{n}A_{i}=A_{n+1}+\sum_{i=m+1}^{n}A_{i}-\sum_{i=m}^{n}A_{i}$$\displaystyle=A_{n+1}+\sum_{i=m+1}^{n}A_{i}-\left( \sum_{i=m+1}^{n}A_{i}\right) -A_{m}=A_{n+1}-A_{m}$.

ou seja, se $a_{i}=A_{i+1}-A_{i}$, então

$\displaystyle\sum_{i=m}^{n}a_{i}=\sum_{i=m}^{n}A_{i+1}-A_{i}=A_{n+1}-A_{m}$

ou

$\displaystyle\sum_{i=m}^{n-1}a_{i}=\sum_{i=m}^{n-1}A_{i+1}-A_{i}=A_{n}-A_{m}$;

Quando $A_{i}$ é uma sucessão crescente, a diferença é positiva. Se $a_{i}=A_{i}-A_{i+1}$, tem-se

$\displaystyle\sum_{i=m}^{n}a_{i}=\sum_{i=m}^{n}A_{i}-A_{i+1}=A_{m}-A_{n+1}$.

ou

$\displaystyle\sum_{i=m}^{n-1}a_{i}=\sum_{i=m}^{n-1}A_{i}-A_{i+1}=A_{m}-A_{n}$.

Já quando $A_{i}$ é decrescente, a diferença é positiva.
$\bigskip$

Exemplos: Calcular a soma

$\displaystyle\sum_{i=1}^{n}a_{i}$,

sendo $a_{i}=A_{i}-A_{i+1}$ e $A_{i}$ dado por

$\bigskip$
(a) $i^{-1};$

$\bigskip$

(b) $i;$

$\bigskip$

(c) $i^{-2};$

$\bigskip$

(d) $i^{2}.$

$\bigskip$

$\blacktriangleright$ Aplicamos as ideias acabadas de expor.

(a)  Como

$a_{i}=A_{i}-A_{i+1}$ $=\displaystyle\frac{1}{i}-\displaystyle\frac{1}{i+1}$ $=\displaystyle\frac{1}{i\left(i+1\right)}$,

vem

$\displaystyle\sum_{i=1}^{n}a_{i}=\displaystyle\sum_{i=1}^{n}\displaystyle\frac{1}{i\left( i+1\right) }$ $=\displaystyle\sum_{i=1}^{n}A_{i}-A_{i+1}=A_{1}-A_{n+1}$

$=\displaystyle\frac{1}{1}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{n}{n+1}$

Logo,

$\displaystyle\sum_{i=1}^{n}$ $\displaystyle\frac{1}{i\left( i+1\right) }$ $=\displaystyle\frac{n}{n+1}$.

$\bigskip$

(b) Neste caso, tem-se

$a_{i}=A_{i}-A_{i+1}$ $= i-\left( i+1\right) =-1$

e

$\displaystyle\sum_{i=1}^{n}a_{i}=$ $\displaystyle\sum_{i=1}^{n}A_{i}-A_{i+1}$ $=\displaystyle\sum_{i=1}^{n}i-\left(i+1\right) =\displaystyle\sum_{i=1}^{n}-1=A_{1}-A_{n+1}=1-\left( n+1\right) =-n$

o que dá o resultado evidente

$\displaystyle\sum_{i=1}^{n}-1=-n$.

$\bigskip$

(c) Agora,

$\displaystyle a_{i}=A_{i}-A_{i+1}$ $=\displaystyle\frac{1}{i^{2}}-\displaystyle\frac{1}{\left( i+1\right) ^{2}}=\displaystyle\frac{2i+1}{i^{2}\left( i+1\right) ^{2}}$

e

$\displaystyle\sum_{i=1}^{n}a_{i}=$  $=\displaystyle\sum_{i=1}^{n}\displaystyle\frac{1}{i^{2}\left( i+1\right) ^{2}}$

$=\displaystyle\sum_{i=1}^{n}A_{i}-A_{i+1}=A_{1}-A_{n+1}=\displaystyle\frac{1}{1^{2}}-\displaystyle\frac{1}{\left( n+1\right) ^{2}}=\displaystyle\frac{n^{2}+2n}{\left(n+1\right) ^{2}}$

donde

$\displaystyle\sum_{i=1}^{n}$ $\displaystyle\frac{2i+1}{i^{2}\left( i+1\right) ^{2}}=\displaystyle\frac{n^{2}+2n}{\left( n+1\right) ^{2}}$

$\bigskip$

(d) Tem-se

$a_i=A_{i}-A_{i+1}=i^2-(i+1)^2=i^2-i^2-2i-1=-2i-1$

e

$\displaystyle\sum_{i=1}^{n}a_i$ $=\displaystyle\sum_{i=1}^{n}A_{i}-A_{i+1}$ $=\displaystyle\sum_{i=1}^{n}i^{2}-\left( i+1\right)^{2}$ $=\displaystyle\sum_{i=1}^{n}-\left( 2i+1\right)=A_{1}-A_{n+1}$ $=1^{2}-\left( n+1\right) ^{2}=-n^{2}-2n$,

novamente evidente, porque, pelo problema 5 , $\displaystyle\sum_{i=1}^{n}i=\dfrac{n(n+1)}{2}$, e $\displaystyle\sum_{i=1}^{n}1=n$.

Por conseguinte,

$\displaystyle\sum_{i=1}^{n}-(2i+1)=-n(n+1)$. $\blacktriangleleft$

Anúncios

Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Caderno, Combinatória, Matemática, Teorema / Teoria com as etiquetas , . ligação permanente.

This site uses Akismet to reduce spam. Learn how your comment data is processed.