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!