7. $\sum_{k=0}^{201}(10 k+7)^{k+1}=7^{1}+17^{2}+27^{3}+37^{4}+\cdots+2017^{202}$ The last digit of the sum is
Solve
$7^{1}+7^{2}+7^{3}+7^{4} \equiv 0(\bmod 10)$, so $\sum_{k=0}^{201}(10 k+7)^{k+1} \equiv 7^{1}+7^{2} \equiv 6(\bmod 10)$, thus the last digit of $\sum_{k=0}^{201}(10 k+7)^{k+1}$ is 6.