`T_1 = 1+1/2`

`T_2 = 1+3/4`

`T_3 = 1+7/8`

`T_4 = 1+15/16`

`T_5 = 1+31/32`

Every term has a '1' in its expression. So the nth term will also have '1' as the beginning value.

When we consider the part after '1+'......

`T_1=>1/2`

`T_2=>3/4`

`T_3=>7/8`

etc....

We can see numerator is 1 less than denominator.

When we consider denominator;

`T_1=>2 = 2^1`

`T_2=>4 = 2^2`

`T_3=>8 = 2^3`

etc....

So the denominator of the nth term will be `2^n` .

Then the numerator will be `2^n - 1`

Finally the nth term will look like;

`T_n = 1+(2^n-1)/(2^n)`