Write down the first few terms of each of the subsequent sequences.

1. {n+1 / n^{2}}^{∞}_{ n=1}

2. {(-1)n+1 / 2n}^{∞}_{ n=0}

3. {bn}^{ ∞}_{ n=1, }where bn = nth digit of ?

**Solution**

1. {n+1 / n^{2}}^{∞}_{ n=1}

To obtain the first few sequence terms here all we require to do is plug in values of n into the formula given and we'll obtain the sequence terms.

**Note:** Inclusion of the "..." at the end! This is a significant piece of notation as it is the just only thing that tells us that the sequence carries on and doesn't end at the last term.

(2) {(-1)n+1 / 2n}^{∞}_{ n=0}

This one is identical to the first one. The main variation is that this sequence doesn't initiate at n = 1.

{(-1) n+1 / 2n}^{∞}_{ n=0} = {=1, ½ , - ¼ , 1/8, - 1/16, ....}

Note: The terms used in this sequence alternate in signs. Sequences of this type are sometimes known as alternating sequences.

(3) {bn}^{ ∞}_{ n=1, }where bn = nth digit of ?

This sequence is not same from the first two in the sense that it doesn't have a particular formula for each term. Though, it does tell us what every term should be. Every term should be the n^{th} digit of ?. So we know that ? = 3.14159265359....

After that the sequence is,

{3,1, 4,1, 5, 9, 2, 6, 5, 3, 5,...}