Mostre que

(a) \displaystyle\sum_{i=0}^{n}i=\displaystyle\frac{n\left( n+1\right) }{2};

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(b) \displaystyle\sum_{i=1}^{n}\dbinom{i}{1}+2\dbinom{i}{2}=\sum_{i=1}^{n}i^{2};

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(c) \displaystyle\sum_{i=r}^{n}\dbinom{i}{r}=\dbinom{n+1}{r+1};

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(d) e calcule o valor da soma \displaystyle\sum_{i=0}^{n}i^{2}.

Resolução

 (a) Se n for ímpar, a soma pretendida é igual à soma de \left( n+1\right) /2 parcelas complementares i e n-i, equidistantes dos extremos, iguais a i+\left( n-i\right) =n:

\displaystyle\sum_{i=0}^{n}i=\displaystyle\sum_{i=1}^{n}i=\displaystyle\sum_{i=1}^{\left( n+1\right) /2}i+\left(n-i\right)=\displaystyle\sum_{i=1}^{\left( n+1\right) /2}i+\displaystyle\sum_{i=1}^{\left( n+1\right)/2}n-\displaystyle\sum_{i=1}^{\left( n+1\right) /2}i=\displaystyle\sum_{i=1}^{\left( n+1\right) /2}n=n\displaystyle\sum_{i=1}^{\left( n+1\right)/2}1=n\times \displaystyle\frac{n+1}{2}.

Se for par, há n/2 parcelas de valor n mais a parcela central n/2:

\displaystyle\sum_{i=0}^{n}i=\left( \displaystyle\sum_{i=1}^{n/2}i+\left( n-i\right) \right)+\displaystyle\frac{n}{2}=\left( \displaystyle\sum_{i=1}^{n/2}i+\displaystyle\sum_{i=1}^{n/2}n-\displaystyle\sum_{i=1}^{n/2}i\right) +\displaystyle\frac{n}{2}=\displaystyle\frac{n}{2}+\sum_{i=0}^{n/2}n=\displaystyle\frac{n}{2}+n\displaystyle\sum_{i=0}^{n/2}1=\displaystyle\frac{n}{2}+n\times \displaystyle\frac{n}{2}=\left( n+1\right) \times \displaystyle\frac{n}{2}

Logo,

\displaystyle\sum_{i=0}^{n}i=\displaystyle\sum_{i=1}^{n}i=\displaystyle\frac{n\left( n+1\right) }{2}.

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(b) De

\dbinom{i}{1}+2\dbinom{i}{2}=i+2\displaystyle\frac{i\left( i-1\right) }{2}=i+i^{2}-i=i^{2},

somando em i, resulta a identidade apresentada.

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(c) Pelo método de indução: para n=1, 0\leq r\leq n=1, tem-se: se r=1,

 \displaystyle\sum_{i=r}^{n}\dbinom{i}{r}=\displaystyle\sum_{i=1}^{1}\dbinom{i}{1}=\dbinom{1}{1}=1

e

 \dbinom{n+1}{r+1}=\dbinom{2}{2}=1;

 se r=0,

\displaystyle\sum_{i=r}^{n}\dbinom{i}{r}=\displaystyle\sum_{i=0}^{1}\dbinom{i}{0}=\dbinom{1}{1}+\dbinom{0}{1}=\dbinom{1}{1}+0=1

e

 \dbinom{n+1}{r+1}=\dbinom{1}{0}=1;

 donde se vê que a identidade se verifica para n=1. Por outro lado, a validade da identidade para n implica

 \displaystyle\sum_{i=0}^{n+1}\dbinom{i}{r}=\left( \displaystyle\sum_{i=0}^{n}\dbinom{i}{r}\right) +\dbinom{n+1}{r}=\dbinom{n+1}{r+1}+\dbinom{n+1}{r}=\dbinom{n+2}{r+1},

 pela identidade de Pascal. Assim, a identidade, porque é válida igualmente para n+1, fica demonstrada.

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(d) Parte-se da identidade de (b), e utiliza-se o resultado de (c):

 \displaystyle\sum_{i=0}^{n}i^{2}=\displaystyle\sum_{i=0}^{n}\dbinom{i}{1}+2\dbinom{i}{2} =\displaystyle\sum_{i=1}^{n}\dbinom{i}{1}+2\displaystyle\sum_{i=1}^{n}\dbinom{i}{2}

 =\dbinom{n+1}{2}+2\dbinom{n+1}{3}=\displaystyle\frac{\left( n+1\right) n}{2}+\displaystyle\frac{\left(n+1\right) n\left( n-1\right) }{3}=\displaystyle\frac{\left( n+1\right) n\left( 2n+1\right) }{6}. \qquad\blacktriangleleft