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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,640

How do I prove

using Mathematica 9?

Is an inductive proof even possible in Mathematica?

*Last edited by ShivamS (2013-11-18 08:49:44)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Hi;

Mathematica already knows that is true but to do the steps

For the base case.

`1^2==(n*(n + 1) (2 n + 1))/6 /.n->1`

True

For the inductive step:

If that is true then

ought to be true. Subtract 1). from 2).

`((n*(n + 1) (2 n + 1))/6 /. n -> n + 1) - (n*(n + 1) (2 n + 1))/6 // FullSimplify`

(n+1)^2

The LHS is obviously (n+1)^2 so we are done.

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**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**ShivamS****Member**- Registered: 2011-02-07
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The first statement doesn't compute properly...

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

Change the = to ==, I am sorry.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**ShivamS****Member**- Registered: 2011-02-07
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Ok, thanks.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

I have made lots of changes to post #2.

Mathematica knows that sum:

`Sum[k^2, {k, 1, n}]`

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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`Sum[...,{i,1,n}]`

seems to be a bit faster than

`Sum[...,{i,n}]`

Why's that?

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**bobbym****Administrator**- From: Bumpkinland
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I guess because he needs time to figure the lower index.

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**anonimnystefy****Real Member**- From: The Foundation
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But, it's an 0.03s difference.

*Last edited by anonimnystefy (2013-11-18 10:07:16)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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If it is not granularity, then that is probably the amount of time it would take.

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**anonimnystefy****Real Member**- From: The Foundation
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Granularity?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Trying to measure a very small increment with a large measuring stick produces granularity.

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**anonimnystefy****Real Member**- From: The Foundation
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How does that happen in M. Do you have an example?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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I do not think I can. If you only have a 3 ft. stick and you and I both try to measure on inch, the measurements will vary greatly.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I do not think that is the problem.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
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Then I would go with the fact that it has to make one more decision.

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**ShivamS****Member**- Registered: 2011-02-07
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Thanks for fixing it.

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**bobbym****Administrator**- From: Bumpkinland
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Hi;

You are welcome.

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