Exista o formula:
[tex] \frac{1}{k(k + r)}= \frac{1}{r}( \frac{1}{k} - \frac{1}{k + n} ) [/tex]
Pe care o poti verifica:
[tex] \frac{1}{r}( \frac{1}{k} - \frac{1}{k + r} ) = \frac{1}{rk}- \frac{1}{r(k+r)}= \frac{(k+r)-k}{rk(k+r)}= \frac{r}{rk(k+n)}= \frac{1}{k(k+n)} [/tex]
In cazul nostru r = 2:
[tex]S= \frac{1}{1*3} + \frac{1}{3*5} + \frac{1}{5*7} +..+ \frac{1}{89*91} \\
S= \frac{1}{2} ( \frac{1}{1} - \frac{1}{3} )+ \frac{1}{2} ( \frac{1}{3} - \frac{1}{5} )+ \frac{1}{2} ( \frac{1}{5} - \frac{1}{7} )+...+ \frac{1}{2} ( \frac{1}{89} - \frac{1}{91} )\\
S= \frac{1}{2}(( \frac{1}{1} - \frac{1}{3} )+ ( \frac{1}{3} - \frac{1}{5} )+( \frac{1}{5} - \frac{1}{7} )+...+( \frac{1}{89} - \frac{1}{91} )) [/tex]
Dupa cum observi se vor reduce: 1/3 cu -1/3, 1/5 cu -1/5, si tot asa pana la 1/89, si ne vor ramane 2 termeni:
[tex]S =\frac{1}{2}(1- \frac{1}{91} ) = \frac{1}{2} * \frac{90}{91} = \frac{45}{91} [/tex]
