Školjka

Neka je $n.$ član niza $\left(a_n\right)_\mathbb{N}$ jednak $n.$ članu tražene sume. Možemo pisati: $$ \begin{aligned} a_n &= \frac{n^2 + (n+1)^2}{n \cdot (n+1)} \\ &= \frac{n^2}{n(n+1)} + \frac{(n+1)^2}{n(n+1)} \\ &= \frac{n}{n+1} + \frac{n+1}{n} \\ &= \frac{n+1-1}{n+1} + \frac{n+1}{n} \\ &= 1 - \frac{1}{n+1} + \frac{n}{n} + \frac{1}{n} \\ &= 2 + \frac{1}{n} - \frac{1}{n+1}. \\ \end{aligned} $$ Tada je suma $k.$ članova niza jednaka: $$ \begin{aligned} \sum_{i=1}^k a_i &= \left(2 + \frac{1}{1} - \frac{1}{1+1}\right) + \left(2 + \frac{1}{2} - \frac{1}{2+1}\right) + \cdots + \left(2 + \frac{1}{k} - \frac{1}{k+1}\right)\\ &= \overbrace{\left(2 + 2 + \cdots + 2\right)}^{k \text{ puta}} + \frac{1}{1} + \left( - \frac{1}{1+1} + \frac{1}{2}\right) + + \left( - \frac{1}{2+1} + \frac{1}{3}\right) + \cdots + \left( - \frac{1}{k-1+1} + \frac{1}{k}\right) - \frac{1}{k+1} \\ &= 2k + 1 - \frac{1}{k+1}. \end{aligned} $$ Specijalno za $k = 99$ tražena suma iznosi $199 - \frac{1}{100}$.