## Answers

# 1/4+3/8+7/16+15/32+31/64 = sum_(r=1)^5 (2^n-1)/2^(n+1) #

#### Explanation:

If we look at the sequence for the terms in denominator we have:

# {4,8,16,32,64}#

It should be obvious that these numbers are successive powers of

# u_n=2^(n+1) # where#n in {1,2,3,4,5}#

And now we look at the sequence for the terms in the numerator

# {1,3,7,15,31} #

and note that the terms are one less than successive powers of

# u_n=2^n-1 # where#n in {1,2,3,4,5}#

Hence a general term of the series would be:

# u_n=(2^n-1)/2^(n+1)# where#n in {1,2,3,4,5}#

And hence we can write the finite series using sigma notation as

# 1/4+3/8+7/16+15/32+31/64 = sum_(r=1)^5 (2^n-1)/2^(n+1) #

NB: The sum evaluates to

#### Explanation:

Note that:

etc...

so the sum is:

We can also write it as:

This last expression allows us also to calculate the sum, as we have the partial sum of a geometric series: