shadyufog0

2022-09-01

Proving with a sequence

The question is :

Show that if $n$ is a power of $2$, then

$$\sum _{i=0}^{{\mathrm{log}}_{2}n-1}{2}^{i}=n-1\phantom{\rule{thickmathspace}{0ex}}.$$

Tried induction at first and tried to prove it on 2n but nothing came out of it. Then i tried every possible way with a series and i'm close but still can't prove it.

Thanks in advance !

The question is :

Show that if $n$ is a power of $2$, then

$$\sum _{i=0}^{{\mathrm{log}}_{2}n-1}{2}^{i}=n-1\phantom{\rule{thickmathspace}{0ex}}.$$

Tried induction at first and tried to prove it on 2n but nothing came out of it. Then i tried every possible way with a series and i'm close but still can't prove it.

Thanks in advance !

bewagox7

Beginner2022-09-02Added 10 answers

If $n$ is a power of $2$ then you have that $n={2}^{m}$ for some $m\in \mathbb{N}$. Thus

$${\mathrm{log}}_{2}n={\mathrm{log}}_{2}{2}^{m}=m{\mathrm{log}}_{2}2=m$$

and therefore

$$\sum _{i=0}^{({\mathrm{log}}_{2}n)-1}{2}^{i}=\sum _{i=0}^{m-1}{2}^{i}{=}^{\text{Geometric sum}}=\frac{1-{2}^{m-1+1}}{1-2}=\frac{1-{2}^{m}}{-1}={2}^{m}-1$$

but since ${2}^{m}=n$, the last term is equal to $n-1$

$${\mathrm{log}}_{2}n={\mathrm{log}}_{2}{2}^{m}=m{\mathrm{log}}_{2}2=m$$

and therefore

$$\sum _{i=0}^{({\mathrm{log}}_{2}n)-1}{2}^{i}=\sum _{i=0}^{m-1}{2}^{i}{=}^{\text{Geometric sum}}=\frac{1-{2}^{m-1+1}}{1-2}=\frac{1-{2}^{m}}{-1}={2}^{m}-1$$

but since ${2}^{m}=n$, the last term is equal to $n-1$

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