Question

Write the first five terms of the sequence. \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\)

Short answer

The first five terms of the sequence are -1, -1/4, -1/9, -1/16, -1/25

Step-by-step solution

Substitute n=1

Put \(n=1\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{1}=\frac{(-1)^{1(1+1) / 2}}{1^{2}}=\frac{(-1)^{1}}{1}=-1\)

Substitute n=2

Put \(n=2\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{2}=\frac{(-1)^{2(2+1) / 2}}{2^{2}}=\frac{(-1)^{3}}{4}= -\frac{1}{4}\)

Substitute n=3

Put \(n=3\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{3}=\frac{(-1)^{3(3+1) / 2}}{3^{2}}=\frac{(-1)^{3}}{9}=-\frac{1}{9}\)

Substitute n=4

Put \(n=4\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{4}=\frac{(-1)^{4(4+1) / 2}}{4^{2}}=\frac{(-1)^{5}}{16}= -\frac{1}{16}\)

Substitute n=5

Put \(n=5\) into \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\), we have \(a_{5}=\frac{(-1)^{5(5+1) / 2}}{5^{2}}=\frac{(-1)^{5}}{25}= -\frac{1}{25}\)