2. Find all values of $n, n \in N$, for which the sum of the first terms of the sequence $a_{k}=3 k^{2}-3 k+1, \quad k \in N$, is equal to the sum of the first $n$ terms of the sequence $b_{k}=2 k+89, k \in N . \quad(12$ points)
Solution. Note that $a_{k}=3 k^{2}-3 k+1=k^{3}-(k-1)^{3}$, and the sum is $S_{n}=n^{3}$. For the second sequence, $b_{k}=2 k+89=(k+45)^{2}-(k+44)^{2}$, the sum is $S_{n}=(n+45)^{2}-45^{2}=n(n+90)$.

We get the equation $n^{3}=n(n+90) \Rightarrow n^{2}-n-90=0 \Rightarrow n=10$.

Answer: 10.