Find the first five terms of the sequence whose nth term is given. a: (-2)n + 2n \\b: n^2 - n...

Using the concept above, we can list the first five terms as follows.

(a) {eq}-2n + 2n {/eq}

This can be simplified down into 0. So, the first five terms of this sequence are {eq}0,0,0,0,0 {/eq}.

(b) {eq}n^2 - n {/eq}

The first five terms can be obtained as follows.

$$\begin{align} a_1 &= (1)^{2} - 1\\[0.3cm] a_1 &= 1-1\\[0.3cm] a_1 &= 0\\[1.0cm] a_2 &= (2)^{2} -2\\[0.3cm] a_2 &= 4 - 2\\[0.3cm] a_2 &= 2\\[1.0cm] a_3 &= (3)^{2} -3\\[0.3cm] a_3 &= 9 - 3\\[0.3cm] a_3 &= 6\\[1.0cm] a_4 &= (4)^{2} -4\\[0.3cm] a_4 &= 16 - 4\\[0.3cm] a_4 &= 12\\[1.0cm] a_5 &= (5)^{2} -5\\[0.3cm] a_5 &= 25 - 5\\[0.3cm] a_5 &= 20\\[1.0cm] \end{align} $$

Hence, the first five terms are {eq}0,2,6,12,\ \text{and}\ 20 {/eq}.

(c) {eq}|10-n^{2}| {/eq}

$$\begin{align} a_1 &= |10-(1)^{2}|\\[0.3cm] a_1 &= |10-1|\\[0.3cm] a_1 &= 9\\[1.0cm] a_2 &= |10-(2)^{2}|\\[0.3cm] a_2 &= |10-4|\\[0.3cm] a_2 &= 6\\[1.0cm] a_3 &= |10-(3)^{2}|\\[0.3cm] a_3 &= |10-9|\\[0.3cm] a_3 &= 1\\[1.0cm] a_4 &= |10-(4)^{2}|\\[0.3cm] a_4 &= |10-16|\\[0.3cm] a_4 &= |-6|\\[0.3cm] a_4 &= 6\\[1.0cm] a_5 &= |10-(5)^{2}|\\[0.3cm] a_5 &= |10-25|\\[0.3cm] a_5 &= |-15|\\[0.3cm] a_5 &= 15\\[1.0cm] \end{align} $$

Hence, the first five terms of the sequence are {eq}9,6,1,6,\ \text{and}\ 15 {/eq}.