Answer:
The general term for the given sequence is:
[tex]a_n=(-1)^{n+1}\dfrac{n}{(n+1)^2}[/tex]
Step-by-step explanation:
The given series is:
[tex]\dfrac{1}4, - \dfrac{2}9, \dfrac{3}{16}, - \dfrac{4}{25}, ......[/tex]
First of all, let us have a look at the positive and negative sign of the sequence.
2nd, 4th, 6th ..... terms have a negative sign.
For this we can use the following
[tex](-1)^{n+1}[/tex]
i.e. Whenever 'n' is odd, power of (-1) will become even resulting in a positive term for odd terms i.e. (1st, 3rd, 5th ........ terms)
Whenever 'n' is even, power of (-1) will become odd resulting in a negative term for even terms i.e. (2nd, 4th, 6th ..... terms)
Now, let us have a look at the numerator part:
1, 2, 3, 4.....
It is simply [tex]n[/tex].
Now, finally let us have a look at the denominator:
4, 9, 16, 25 ......
There are squares of the (n+1).
i.e. 1st term has a square of 2.
2nd term has a square of 3.
and so on
So, it can be represented as:
[tex](n+1)^2[/tex]
[tex]\therefore[/tex] nth term of the sequence is:
[tex]a_n=(-1)^{n+1}\dfrac{n}{(n+1)^2}[/tex]
