Question

How do you use sigma notation to write the sum for `1/4+3/8+7/16+15/32+31/64`?

Answer 1

`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 `2`, with the first term `2` missing so the general term of the above sequence would be:

`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 `2`, so the general term of this sequence be:

`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 `129/64`

Answer 2

`sum_(n=1)^5 (2^n-1)/(2^(n+1))`

Explanation:

Note that:

`1/4 = (2^1-1)/2^2`

`3/8 = (2^2-1)/2^3`

etc...

so the sum is:

`sum_(n=1)^5 (2^n-1)/(2^(n+1))`

We can also write it as:

`S = sum_(n=1)^5 (2^n-1)/(2^(n+1)) = sum_(n=1)^5 (1/2-1/(2^(n+1)) )= 5/2 -1/4sum_(n=0)^4 (1/2)^n`

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

`sum_(k=0)^n x^k = (1-x^(k+1))/(1-x)`

`S = 5/2 -1/4sum_(n=0)^4 1/2^n = 5/2 - 1/4(1-1/2^5)/(1-1/2) = 5/2- 1/4(31/32)/(1/2) = 5/2- 31/64 = 129/64`