Question

Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. text({) 1/4, - 2/9, 3/16, - 4/25, ...text(})

Answer 1

Answer:

The general term for the given sequence is:

$a_n=(-1)^{n+1}\dfrac{n}{(n+1)^2}$

Step-by-step explanation:

The given series is:

$\dfrac{1}4, - \dfrac{2}9, \dfrac{3}{16}, - \dfrac{4}{25}, ......$

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

$(-1)^{n+1}$

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 $n$.

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:

$(n+1)^2$

$\therefore$ nth term of the sequence is:

$a_n=(-1)^{n+1}\dfrac{n}{(n+1)^2}$