5. Zadatak

Primijetimo da je $n$-ti član danog zbroja oblika

$$a_n=\frac{2n+1}{n^2\cdot(n+1)^2}.$$

Nadalje, član $a_n$ možemo zapisati u sljedećem obliku:

$$a_n=\frac{2n+1}{n^2\cdot(n+1)^2}=\frac n{n^2\cdot(n+1)^2}+\frac{n+1}{n^2\cdot(n+1)^2}$$ $$=\frac1{n\cdot(n+1)^2}+\frac1{n^2\cdot(n+1)}$$ $$=\frac1{n(n+1)}\left(\frac1{n+1}+\frac1n\right)$$ $$=\left(\frac1n-\frac1{n+1}\right)\left(\frac1{n+1}+\frac1n\right)$$ $$=\left(\frac1n\right)^2-\left(\frac1{n+1}\right)^2.$$

Dakle, traženi zbroj jednak je

$$\frac3{1^2\cdot2^2}+\frac5{2^2\cdot3^2}+\frac7{3^2\cdot4^2}+\cdots+\frac{199}{99^2\cdot100^2}$$ $$=\left[1-\left(\frac12\right)^2\right]+\left[\left(\frac12\right)^2-\left(\frac13\right)^2\right]+\left[\left(\frac13\right)^2-\left(\frac14\right)^2\right]+\cdots+\left[\left(\frac1{99}\right)^2-\left(\frac1{100}\right)^2\right]$$ $$=1-\left(\frac1{100}\right)^2=1-\frac1{10000}=\frac{9999}{10000}.$$