## Precalculus (10th Edition)

$1,3,6,10,15,21,28$
We are given: $u_1=1$ Compute $u_n$ by substituting $1, 2, 3, 4, 5, 6$ for $n$ in the formula: $u_{n+1}=u_n+(n+1)$ $u_2=u_1+(1+1)=1+2=3$ $u_3=u_2+(2+1)=3+3=6$ $u_4=u_3+(3+1)=6+4=10$ $u_5=u_4+(4+1)=10+5=15$ $u_6=u_5+(5+1)=15+6=21$ $u_7=u_6+(6+1)=21+7=28$