## Somas

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}$. $\qquad\blacktriangleleft$