1. Zadatak

Prvo rješenje.

Dokaz provodimo matematičkom indukcijom.

Baza indukcije – provjeravamo da formula vrijedi za $n=2$:

$$\sqrt{1+\frac1{1^2}+\frac1{2^2}}=\sqrt{\frac94}=\frac32=2-\frac12.$$

Pretpostavimo da formula vrijedi za neki prirodni broj $n\geq2$.

Korak indukcije – koristeći pretpostavku treba pokazati da formula vrijedi i za $n+1$, tj. da vrijedi:

$$\sqrt{1+\frac1{1^2}+\frac1{2^2}}+\sqrt{1+\frac1{2^2}+\frac1{3^2}}+\cdots+ \sqrt{1+\frac1{(n-1)^2}+\frac1{n^2}}+ \sqrt{1+\frac1{n^2}+\frac1{(n+1)^2}}=n+1-\frac1{n+1}.$$

Lijeva strana je jednaka (po pretpostavci):

$$\begin{aligned} n-\frac1n+\sqrt{\frac{n^2(n+1)^2+(n+1)^2+n^2}{n^2(n+1)^2}} &=n-\frac1n+\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n(n+1)}\\ &=n-\frac1n+\frac{\sqrt{(n^2+n+1)^2}}{n(n+1)}\\ &=n-\frac1n+\frac{n^2+n+1}{n(n+1)}\\ &=n-\frac1n+\frac{n(n+1)+1}{n(n+1)}\\ &=n-\frac1n+1+\frac1{n(n+1)}\\ &=n+1-\frac{(n+1)-1}{n(n+1)}\\ &=n+1-\frac1{n+1}, \end{aligned}$$

što smo i trebali dobiti.

Drugo rješenje.

Izraz pod korijenom jednak je:

$$\begin{aligned} 1+\frac1{(n-1)^2}+\frac1{n^2} &=\frac{(n-1)^2n^2+n^2+(n-1)^2}{(n-1)^2n^2}\\ &=\frac{n^4-2n^3+n^2+n^2+n^2-2n+1}{(n-1)^2n^2}\\ &=\frac{n^4+n^2+1-2n^3+2n^2-2n}{(n-1)^2n^2}\\ &=\frac{(n^2-n+1)^2}{(n-1)^2n^2}. \end{aligned}$$

Svaki pribrojnik je oblika:

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

Konačna suma iznosi:

$$\begin{aligned} &\left(1+\frac11-\frac12\right)+ \left(1+\frac12-\frac13\right)+\cdots+ \left(1+\frac1{n-1}-\frac1n\right)\\ &=(n-1)\cdot1+ \left(\frac11-\frac12+\frac12-\frac13+\cdots+\frac1{n-1}-\frac1n\right)\\ &=n-1+\frac11-\frac1n=n-\frac1n. \end{aligned}$$