parmenides51 wrote: It is known that the sequence $\{a_n\}$ satisfies $a_1=2$, $a_n=2^{2n}a_{n-1}+n\cdot 2^{n^2}$, $(n \ge 2)$, find the general term of $a_n$. Let $b_n=\frac{a_n}{2^{n^2+n}}$, then $b_n=b_{n-1}+\frac{n}{2^n}$. So $b_n=b_1+\sum_{k=2}^{n} \frac{k}{2^k} =\sum_{k=1}^{n} \frac{k}{2^k}=\frac{2^{n+1}-n-2}{2^n}$ $\implies$ $a_n=2^{n^2}(2^{n+1}-n-2)$