Write the first five terms of each of the following sequences whose nth terms are:
$ a_{n}=\frac{n(n-2)}{2} $

Given:

\( a_{n}=\frac{n(n-2)}{2} \)

To do:

We have to find the first five terms of the given sequence.

Solution:

$a_n=\frac{n(n-2)}{2}$

Taking $n=1$, we get

$a_1=\frac{n(n-2)}{2}=\frac{1(1-2)}{2}=\frac{1(-1)}{2}=\frac{-1}{2}$

Taking $n=2$, we get

$a_2=\frac{n(n-2)}{2}=\frac{2(2-2)}{2}=\frac{2(0)}{2}=0$

Taking $n=3$, we get

$a_3=\frac{n(n-2)}{2}=\frac{3(3-2)}{2}=\frac{3(1)}{2}=\frac{3}{2}$

Taking $n=4$, we get

$a_4=\frac{n(n-2)}{2}=\frac{4(4-2)}{2}=\frac{4(2)}{2}=4$

Taking $n=5$, we get

$a_5=\frac{n(n-2)}{2}=\frac{5(5-2)}{2}=\frac{5(3)}{2}=\frac{15}{2}$

Hence, the first five terms of the given sequence are $\frac{-1}{2}, 0, \frac{3}{2}, 4$ and $\frac{15}{2}$.  

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Updated on: 10-Oct-2022

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