`1 + 1/2, 1 + 3/4, 1 + 7/8, 1 + 15/16, 1 + 31/32` Write an expression for the apparent nth term of the sequence. (assume that n begins with 1)

Expert Answers
jeew-m eNotes educator| Certified Educator

`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)`

htennis | Student

When we look at the sequence as a whole, we see that there is always a 1 added to a fraction, and this is important. So we try to see if there is any connection between the fractions. 1/2, 3/4, 7/8, etc. We see that the difference between the numerator and the denominator is always 1. But what is the relationship between each denominator? If you take the denominators: 2, 4, 8, 16, etc, you see that they are all powers of 2. In addition, the numerator would be that power of 2 minus 1. 2 is 2^1, 4 is 2^2, 8 is 2^3, etc. Therefore, the sequence for the nth term would be 1+[(2^n-1)/2^n].