6. Zadatak

Uočite da se može zapisati i kao:

$$\frac1{2\cdot4}+\frac1{4\cdot6}+\frac1{6\cdot8}+\cdots+\frac1{98\cdot100}.$$

Svaki razlomak sada ima oblik $\frac1{4k(k+1)}$, gdje je $k=1,2,3,\ldots,49$.

Rastavimo razlomak kao

$$\begin{aligned} \frac1{k(k+1)}&=\frac Ak+\frac B{k+1}=\frac{Ak+A+Bk}{k(k+1)}, \tag{1}\\ A+B&=0,\\ A&=1,\\ B&=-1,\\ \frac1{k(k+1)}&=\frac1k-\frac1{k+1},\\ \frac1{4k(k+1)}&=\frac14\left(\frac1k-\frac1{k+1}\right). \end{aligned}$$

Sada početni zbroj možemo zapisati kao:

$$\frac14\left(1-\frac12\right)+\frac14\left(\frac12-\frac13\right)+\frac14\left(\frac13-\frac14\right)+\cdots+\frac14\left(\frac1{49}-\frac1{50}\right). \tag{2}$$

Izlučimo li $\frac14$, dobit ćemo:

$$\frac14\left(1-\frac1{50}\right)=\frac14\cdot\frac{49}{50}=\frac{49}{200}.$$