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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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