Chapter 11 - 11.2 - Arithmetic Sequences and Partial Sums - 11.2 Exercises - Page 786: 16

See below.

Plugging in the different values: $a_1=1+4(1-1)=1+4(0)=1+0=1$ $a_2=1+4(2-1)=1+4(1)=1+4=5$ $a_3=1+4(3-1)=1+4(2)=1+8=9$ $a_4=1+4(4-1)=1+4(3)=1+12=13$ $a_5=1+4(5-1)=1+4(4)=1+16=17$ $a_5-a_4=a_4-a_3=a_3-a_2=a_2-a_1=4$, thus this is an arithmetic sequence with a common difference of $4$.