Another Form For A Sum Of Geometric progression

Question

I have the following summation: $$\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{22}}$$

Which is the sum of the $23$ elements of a $32$ bit mantissa.

Now the sum is:

$$\frac{\frac{1}{2}(\frac{1}{2^{23}}-1)}{\frac{1}{2}-1}=-(2^{-23}-1)=1-2^{-23}$$

In the note it was calculate by $$(2^{23}-1)2^{-23}$$ which is the same, but how did we get to this formula this from the first place?

Answer 1

Considering the expression $1-2^{-23}$, multiply and divide by $2^{-23}$. So, you obtain: $$\frac{1-2^{-23}}{2^{-23}}\cdot2^{-23}=(\frac{1}{2^{-23}}-1)\cdot2^{-23}=(2^{23}-1)\cdot2^{-23}$$

Answer 2

The sum is $1-\frac{1}{2^{22}}$ $$$$ Denote that $$T=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{22}}$$ Then we have $$2T=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{21}}$$ Hence,$$T=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{21}}-(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{22}})=1-\frac{1}{2^{22}}$$

Answer 3

let $s=1+\frac1r+\frac1{r^2}...eq-1$ multiply with r on both sides we get $r*s=r+1+\frac1r+\frac1{r^2}...eq-2$ substituting eq-1 in eq-2 we get $r*s=r+s==>s=\frac{r}{r-1}$ which can be written as $\frac1{1-{\frac1r}}$ where s:-is sum & r:-is common ratio.from above basic derivation you can get your formula.