# How do you write the first six terms of the sequence a_n=(n-1)^2?

Oct 10, 2016

First six terms are $\left\{0 , 1 , 4 , 9 , 16 , 25\right\}$

#### Explanation:

You can derive the ${n}^{t h}$ term by puuting the value of $n$ in ${a}_{n} = {\left(n - 1\right)}^{2}$

Hence, ${a}_{1} = {\left(1 - 1\right)}^{2} = 0$

${a}_{2} = {\left(2 - 1\right)}^{2} = {1}^{2} = 1$

${a}_{3} = {\left(3 - 1\right)}^{2} = {2}^{2} = 4$

${a}_{4} = {\left(4 - 1\right)}^{2} = {3}^{2} = 9$

${a}_{5} = {\left(5 - 1\right)}^{2} = {4}^{2} = 16$

${a}_{6} = {\left(6 - 1\right)}^{2} = {5}^{2} = 25$

Hence, first six terms are $\left\{0 , 1 , 4 , 9 , 16 , 25\right\}$