Find the summation formula of?

Question

I am trying to represent the following in summation form,

Ex 1: let say the upper bound is 16, the lower bound is 1

The summation should be able to give the sum of 16 + 8 + 4 + 2 + 1

let say the upper bound is 64, the lower bound is 1

Ex 2: The summation should be able to give the sum of 64 + 32 + 16 + 8 + 4 + 2 + 1

Here is what I have came up so far:

$$\sum_{i=1}^n i/2 $$

Can anyone tell me if I am on the right track or not?

Answer 1

The thing you are trying to do is not how you use the summation notation.

When you write something using the notation $$\sum_{i=1}^8,$$ the sum will have $8$ terms: one for $i=1,$ one for $i=2,$ one for $i=3,$ and so forth. The last two terms will be the terms for $i=7$ and $i=8.$

If you want to "skip" values of $i$ you must use a different notation.

The usual way to get around this is to notice how many terms you actually want, and find a way so that when you plug $i$ into the formula to the right of the $\sum,$ you get the $i$th term. For example, for $1+2+4+8+16+32+64,$ we see there are $7$ terms in the sum, so we can start by writing $$\sum_{i=1}^7.$$

But now we have to write something for what each term will be. Notice that when $i=1$ we want the term to be $1,$ when $i=2$ we want the term to be $2,$ but when $i=3$ we want the term to be $4$ (not $3$), when $i=4$ we want $8,$ and so forth, doubling each time. "Doubling each time" means a power of $2$ is involved. A little trial and error will show that $2^{i-1}$ will work: \begin{align} \sum_{i=1}^7 2^{i-1} &= 2^{1-1} + 2^{2-1} + 2^{3-1} + 2^{4-1} + 2^{5-1} + 2^{6-1} + 2^{7-1}\\ &= 1 + 2+4+8+16+32+64. \end{align}

Answer 2

Since you are adding power of two from largest to smallest, the summation should be written as

$$\sum_{r=0}^n 2^{n-r}$$ where $U=2^n \;\; (n\in\mathbb Z) $ being the upper bound.