Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set - Page 894: 32

$a_{1}=(-1)^1(1)^2=-1$ $a_{2}=(-1)^2(2)^2=4$ $a_{3}=(-1)^3(3)^2=-9$ $a_{4}=(-1)^4(4)^2=16$ $a_{10}=(-1)^{10}(10)^2=100$ $a_{15}=(-1)^{15}(15)^2=-225$

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(-1)^1(1)^2=-1$ $a_{2}=(-1)^2(2)^2=4$ $a_{3}=(-1)^3(3)^2=-9$ $a_{4}=(-1)^4(4)^2=16$ $a_{10}=(-1)^{10}(10)^2=100$ $a_{15}=(-1)^{15}(15)^2=-225$