# Sequences

**Question**

Find in terms of n, the n th term of the sequence:

1/3 , 2/5 , 3/7, 4/9 , 5/11 , . . .

**Answer**

Here is a neat way to look at it…

Lets count for n = 1,2,3 etc. As we “count” we “observe” how this counting “relates” to the “pattern of the numbers given”.

Consider the “numerator” and “denominator” pattern of numbers separately.

For the “numerator”, we have :

n = 1 ,Numerator value : 1

n = 2 ,Numerator value : 2

n = 3 ,Numerator value : 3

n = 4 ,Numerator value : 4

n = 5 ,Numerator value : 5

See the pattern?

…n = m, Numerator Value : m.

Now consider the “denominator” values in the similar manner.

n = 1 ,Denominator value : 3 = 2(1) + 1

n = 2 ,Denominator value : 5 = 2(2) + 1

n = 3 ,Denominator value : 7 = 2(3) +1

n = 4 ,Denominator value : 9 = 2(4) +1

n = 5 ,Denominator value : 11 = 2(5) +1

See the pattern?

…n = m, Numerator Value : 2(m) + 1.

Hence “combining” our observations above,

The “nth term” will be = (n)/[ 2(n) + 1]

That’s it!