HOW TO PROVE A FORMULA FOR THE SUM OF POWERS OF 2 BY INDUCTION?

Here is where I"m getting off traông chồng. Lets look at the right side of the last equation: $2^n+1 -1$ I can rewrite this as the following.

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$2^1(2^n) - 1$ But, from our hypothesis $2^n = 2^n+1 - 1$ Thus:

$2^1(2^n+1 -1) -1$ This is where I get lost. Because when I distribute through I get this.

$2^n+2 -2 -1$ This is wrong is it not? Am I not applying the rules of exponents correctly here? I have sầu the solution so I know what I"m doing is wrong. Here is the correct proof.

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summation induction
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edited Mar 8 "15 at 4:14
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asked Feb 18 "11 at 0:37
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An easy way lớn do this is using binary. Here"s an idea of what I mean:

$2^0$ in binary is $1$$2^1$ in binary is $10$$2^2$ in binary is $100$

For a general rule:

$2^n$ in binary is $100cdots0$ (n zeros)

Add those together và you get $2^0 + 2^1 + ... + 2^n$ in binary is $11...11$ ($n+1$ ones).

Now it"s obvious that adding 1 khổng lồ that gives you$$100cdots00 quad ext ($n+1$ zeros)$$Which as we all know is $2^n+1$.

Thus $2^n+1 - 1$ is equal so the sum of powers of two up lớn $2^n$.


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edited Jan 11 "16 at 0:30
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JnxF
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answered Jan 11 "16 at 0:12
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Both

Inductive Step lớn prove sầu is: $ 2^n+1 = 2^n+2 - 1$Our hypothesis is: $2^n = 2^n+1 -1$

are wrong và should be

Inductive Step to prove is: $ 2^0 + 2^1 + ... + 2^n + 2^n+1 = 2^n+2 - 1$Our hypothesis is: $ 2^0 + 2^1 + ... + 2^n = 2^n+1-1$

Add $2^n+1$ to lớn both sides of the hypothesis & you have the step lớn prove since $2^n+1-1 +2^n+1 = 2^n+2 - 1$


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answered Feb 18 "11 at 0:44
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HenryHenry
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HINT $ $ Here"s the inductive sầu proof for summing a general geometric series.

THEOREM $ mquad 1 + x + cdots + x^n-1 = dfracx^n-1x-1$

Proof $ $ Base case: It is true for $ m n = 1:,:$ viz. $ m 1 = (x-1)/(x-1):$.

Inductive step: Suppose it is true for $ m n = k:. $ Then we have

$$ m x^k + (x^k-1 +: cdots: + 1): =: x^k +fracx^k-1x-1 = fracx^k+1-1x-1$$

which implies it is true for $ m: n = k+1:,:$ thus completing the inductive sầu proof.

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The proof you seek is just the special case $ m x = 2 $.


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answered Feb 18 "11 at 1:19
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Bill DubuqueBill Dubuque
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I don"t see the answer I like here, so I"m writing my own.

Basic proof:

We wish to lớn prove $2^0 + 2^1 + ... + 2^n-1 = 2^n - 1$ for all $n$. We can verify by inspection this is true for n=1. Next, assume that $2^0 + 2^1 + ... + 2^n = 2^n+1 - 1$.

$(2^0 + 2^1 + ... + 2^n) + 2^n+1 = (2^n+1 - 1) + 2^n+1 = 2 cdot 2^n+1 - 1 = 2^n+2 - 1$, so we have sầu shown $2^0 + 2^1 + ... + 2^n-1 = 2^n - 1$ is true for all n.


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edited Jan 18 at 7:52
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answered Atruyền thông quảng cáo 9 "18 at 4:12
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