Chapter 8, Sequences and Series - Chapter 8 Review - Exercises - Page 639: 7

$a_1=1$ $a_2=4$ $a_3=9$ $a_4=16$ $a_{5}=25$ $a_{6}=36$ $a_{7}=49$

We are given: $a_{n}=a_{n-1}+2n-1$ and $a_1=1$ We evaluate: $a_1=1$ $a_2=a_1+4-1=1+4-1=4$ $a_3=a_2+6-1=4+6-1=9$ $a_4=a_3+8-1=9+8-1=16$ $a_{5}=a_{4}+10-1=16+10-1=25$ $a_{6}=a_{5}+12-1=25+12-1=36$ $a_{7}=a_{6}+14-1=36+14-1=49$