Skaitļu virknei \(\left(a_{i}\right)\) visiem \(n > 1\) ir spēkā sakarība \(a_{1}+a_{2}+\ldots+a_{n}=n^{2} a_{n}\). Aprēķināt \(a_{50}\), ja zināms, ka \(a_{1}=1000\).
Apzīmējam \(S_{n}=a_{1}+a_{2}+\ldots+a_{n}\). Tad \(n^2 a_n=S_n\) , \((n+1)^{2}a_{n+1}=S_{n+1}\) un
\[\begin{gathered} S_{n+1}-S_{n}=a_{n+1}=(n+1)^{2}a_{n+1}-n^{2}a_{n} \Rightarrow a_{n+1}\left(n^{2}+2n\right)=n^{2}a_{n} \Rightarrow \\ a_{n+1}=\frac{n}{n+2} \cdot a_{n}=\frac{n}{n+2} \cdot \frac{n-1}{n+1} a_{n-1}=\frac{n}{n+2} \cdot \frac{n-1}{n+1} \cdot \frac{n-2}{n} a_{n-2}=\ldots=\frac{1}{n+2} \cdot \frac{1}{n+1} \cdot 2 \cdot 1 \cdot a_{1} \end{gathered}\]
Esam ieguvuši, ka \(a_{n}=\frac{2 a_{1}}{(n+1) n}\) un varam aprēķināt prasīto \(a_{50}=\frac{2 \cdot 1000}{51 \cdot 50}=\frac{40}{51}\).