MHB Explaining Integer Equations: Why -1 at the End?

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In integer equations that start with 1, the presence of -1 at the end arises from the manipulation of the equation. When the sum is expressed as S = 1 + 2 + 4 + ... + 2^N, subtracting 1 from both sides leads to a new equation where the original "+1" becomes "-1" on the left side. This transformation is essential for deriving the formula S = 2^(N+1) - 1. The -1 reflects the initial "+1" in the sum, which balances the equation after manipulation. Understanding this concept clarifies why the -1 appears in specific cases but not when starting with other integers like -5 or -8.
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Can some one explain to my why an integer equation that starts with 1 has a -1 at the end of the equation.

example:

1 + 2 + 4 + 8 + 16 ... + 2 ^ N = 2 x ( 2 ^ N ) - 1

Conceptually where does the rule come from that there is a minus at the end of the equation.

It starts with an odd number so the answer must be an odd number and that's why -1 is subtracted at the end then how come if an equation started with -5 or -8 you would not subtract the -5 or -8 in the end?

I've seen numerical explanations but they confuse me, can it be explained with words?
 
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Re: Intiger Concepts

psc109 said:
Can some one explain to my why an integer equation that starts with 1 has a -1 at the end of the equation.

example:

1 + 2 + 4 + 8 + 16 ... + 2 ^ N = 2 x ( 2 ^ N ) - 1

Conceptually where does the rule come from that there is a minus at the end of the equation.

It starts with an odd number so the answer must be an odd number and that's why -1 is subtracted at the end then how come if an equation started with -5 or -8 you would not subtract the -5 or -8 in the end?

I've seen numerical explanations but they confuse me, can it be explained with words?

The general formula for a geometric sum is...

$\displaystyle S_{n} = \sum_{k=0}^{n} a^{k} = \frac{1 - a^{n+1}}{1-a}\ (1)$

Setting a=2 You obtain $S_{n} = 2^{\ n+1} - 1$...

Kind regards

$\chi$ $\sigma$
 
Re: Intiger Concepts

Thank you Chisigma, the example you gave does not start with one and does not have a negative 1 at the end.

Why is it there in the the example i was given? Why subtract 1 at the end when if say I started it 8 I would not subtract 8 at the end.
 
Re: Intiger Concepts

psc109 said:
Thank you Chisigma, the example you gave does not start with one and does not have a negative 1 at the end.

Why is it there in the the example i was given? Why subtract 1 at the end when if say I started it 8 I would not subtract 8 at the end.

I confess not to understand exactly Your question, expecially the words 'it start with 1 and ends with -1'... ewerywhere in Math we meet expressions like 'a = b' and no general connection exists between the 'head' of a and the 'tail' of b... may be however that my misundestanding depends from my poor knowledege of the english language (Emo)...

Kind regards $\chi$ $\sigma$
 
One way of looking at it is this: Let S= 1+ 2+ 2^2+ 2^3+ ... 2^n
Subtract 1 from both sides: S- 1= 2+ 2^2+ 2^3+ ...+ 2^n.

The "+1" on the right has become "-1" on left!

Now, factor a "2" on the right: S- 1= 2(1+ 2+ 2^2+ ...+ 2^{n-1})

That "1+ 2+ 2^2+ 2^{n-1} on the right is almost the "1+ 2+ 2^2+ ...+ 2^n" we had before. Make it that way by "adding and subtracting 2^n inside the parentheses:
S-1= 2(1+ 2+ 2^2+ ...+ 2^{n-1}+ 2^n- 2^n)= 2(S- 2^n)
S- 1= 2S- 2^{n+1}

Solve that for S by subtracting 2S from both sides and adding 1 to both sides:
-S= 1- 2^{n+1}

so S= 2^{n+1}- 1

That "-1" at the end is due to the "1" on the left in "S- 1" which is, itself, due to the fact that there was a "+1" on the right in the initial equation.
 
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