## Precalculus (6th Edition) Blitzer

Given $a_j=j^2+1$, we have: $a_1=1^2+1=2$ $a_2=2^2+1=5$ $a_3=3^2+1=10$ $a_4=4^2+1=17$ $a_5=5^2+1=26$ $a_6=6^2+1=37$ The sum is $\sum_1^6 a_j=2+5+10+17+26+37=97$