Emanuel Keith

2022-06-28

Since everyone freaked out, I made the variables are the same.

$\sum _{x=1}^{n}{2}^{x-1}$

$\sum _{x=1}^{n}{2}^{x-1}$

Ryan Newman

Beginner2022-06-29Added 26 answers

In this instance, without explicitly using the formula for geometric series,

$\begin{array}{rl}\sum _{x=1}^{n}{2}^{x-1}& =1+2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1}\\ & =1+(1+2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1& \text{add and subtract}\text{}1\\ & =(1+1)+(2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1& \text{regroup}\\ & =2+(2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1\\ & =(2+2)+({2}^{2}+\cdots +{2}^{n-1})-1& \text{regroup again}\\ & ={2}^{2}+({2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1\\ & =({2}^{2}+{2}^{2})+({2}^{3}+\cdots +{2}^{n-1})-1& \text{regroup again}\\ & ={2}^{3}+({2}^{3}+\cdots +{2}^{n-1})-1\\ & =\cdots & \text{lather, rinse, repeat}\\ & ={2}^{n-1}+({2}^{n-1})-1& \text{nearly done}\\ & ={2}^{n}-1.\end{array}$

Now that we know the form of the result, it is also possible to prove the result

$\sum _{x=1}^{n}{2}^{x-1}={2}^{n}-1$

more formally by induction. Clearly, the result holds when n=1 since ${2}^{0}={2}^{1}-1$. Then, if the result holds for some positive integer n, we have that

$\sum _{x=1}^{n+1}{2}^{x-1}=\sum _{x=1}^{n}{2}^{x-1}+{2}^{n}=({2}^{n}-1)+{2}^{n}={2}^{n+1}-1$

and so the result holds for n+1 as well. Since we know that the result holds when n=1, it follows by induction that it holds for all positive integers n.

$\begin{array}{rl}\sum _{x=1}^{n}{2}^{x-1}& =1+2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1}\\ & =1+(1+2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1& \text{add and subtract}\text{}1\\ & =(1+1)+(2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1& \text{regroup}\\ & =2+(2+{2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1\\ & =(2+2)+({2}^{2}+\cdots +{2}^{n-1})-1& \text{regroup again}\\ & ={2}^{2}+({2}^{2}+{2}^{3}+\cdots +{2}^{n-1})-1\\ & =({2}^{2}+{2}^{2})+({2}^{3}+\cdots +{2}^{n-1})-1& \text{regroup again}\\ & ={2}^{3}+({2}^{3}+\cdots +{2}^{n-1})-1\\ & =\cdots & \text{lather, rinse, repeat}\\ & ={2}^{n-1}+({2}^{n-1})-1& \text{nearly done}\\ & ={2}^{n}-1.\end{array}$

Now that we know the form of the result, it is also possible to prove the result

$\sum _{x=1}^{n}{2}^{x-1}={2}^{n}-1$

more formally by induction. Clearly, the result holds when n=1 since ${2}^{0}={2}^{1}-1$. Then, if the result holds for some positive integer n, we have that

$\sum _{x=1}^{n+1}{2}^{x-1}=\sum _{x=1}^{n}{2}^{x-1}+{2}^{n}=({2}^{n}-1)+{2}^{n}={2}^{n+1}-1$

and so the result holds for n+1 as well. Since we know that the result holds when n=1, it follows by induction that it holds for all positive integers n.

Cory Patrick

Beginner2022-06-30Added 6 answers

You're saying you want as outputs

$1,3,7,15,31,63$

Note they are respectively ${2}^{1}-1,{2}^{2}-1,{2}^{3}-1,{2}^{4}-1,{2}^{5}-1,{2}^{6}-1$ so what you really want is

$f(n)={2}^{n}-1$

Now this is a finite geometric sum, namely

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

Now this is a finite geometric sum, namely

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

This follows from the geometric sum formula, that is

$\sum _{i=0}^{n-1}{a}^{i}=\frac{{a}^{n}-1}{a-1}$

The MO for this is the following. Let our sum be S

$1+a+\cdots +{a}^{n-1}=S$

Then

$a+{a}^{2}+\cdots +{a}^{n}=aS$

But

$a+{a}^{2}+\cdots +{a}^{n}=\left(1+a+\cdots +{a}^{n-1}\right)-1+{a}^{n}=S-1+{a}^{n}$

So that

$\begin{array}{rl}& S-1+{a}^{n}=aS\\ & S-aS=1-{a}^{n}\\ & \left(1-a\right)S=1-{a}^{n}\\ & S=\frac{1-{a}^{n}}{1-a}\end{array}$

as desired.

$1,3,7,15,31,63$

Note they are respectively ${2}^{1}-1,{2}^{2}-1,{2}^{3}-1,{2}^{4}-1,{2}^{5}-1,{2}^{6}-1$ so what you really want is

$f(n)={2}^{n}-1$

Now this is a finite geometric sum, namely

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

Now this is a finite geometric sum, namely

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

$\sum _{i=0}^{n-1}{2}^{i}={2}^{n}-1$

This follows from the geometric sum formula, that is

$\sum _{i=0}^{n-1}{a}^{i}=\frac{{a}^{n}-1}{a-1}$

The MO for this is the following. Let our sum be S

$1+a+\cdots +{a}^{n-1}=S$

Then

$a+{a}^{2}+\cdots +{a}^{n}=aS$

But

$a+{a}^{2}+\cdots +{a}^{n}=\left(1+a+\cdots +{a}^{n-1}\right)-1+{a}^{n}=S-1+{a}^{n}$

So that

$\begin{array}{rl}& S-1+{a}^{n}=aS\\ & S-aS=1-{a}^{n}\\ & \left(1-a\right)S=1-{a}^{n}\\ & S=\frac{1-{a}^{n}}{1-a}\end{array}$

as desired.

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