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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.
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Exemplos: Calcular a soma

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

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

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(a) i^{-1};

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(b) i;

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(c) i^{-2};

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(d) i^{2}.

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\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}.

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(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.

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(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}}

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(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=\frac{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