Question

Find the nth term of the harmonic sequence $-1,\frac{1}{4},\frac{1}{-9},\frac{1}{16}$.

• A. $(-1{)}^n.n^2$
• B. $(-1{)}^n.n$
• C. $\frac{1}{(1{)}^n.n^2}$
• D. $\frac{1}{(-1{)}^n.n^2}$

Answer 1

The correct option is D $\frac{1}{(-1{)}^n.n^2}$

Given series is $-1,\frac{1}{4},\frac{1}{-9},\frac{1}{16}$

In the given series, the denominator is square to the no. of term i.e for the first term it will be $1^2$.

For $n^{th}$ term : $=\frac{1}{n^2}$

But there is alternatively $(-)$ sign also,

So, the $n^{th}$ term will be $=\frac{1}{(-1{)}^n{n}^2}$.