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I use induction N=1 And n=k But in case of n=k+1 i can't write the thing .help

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saulspatz
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Akash Sarkar
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  • The proof becomes easier if you use the formulas for $1+2+3+\cdots+n$ and $1+2^3+3^3+\cdots +n^3$ and prove them via induction. – Peter Feb 03 '19 at 13:25
  • Explain please I can't get answer – Akash Sarkar Feb 03 '19 at 13:34
  • You can have a look at https://en.wikipedia.org/wiki/File:Nicomachus_theorem_3D.svg – Dr. Mathva Feb 03 '19 at 13:34
  • The formula for $$S(n)=1+2+3+\cdots +n$$ can easily be found (even without induction) : You can write the sum in reverse $$S(n)=n+\cdots+3+2+1$$ and immediately see that $$2S(n)=n\cdot(n+1)$$ Now show by induction that $$1+2^3+3^3+\cdots +n^3=S(n)^2$$ holds for all $n$. – Peter Feb 03 '19 at 13:38
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    @Peter thanks ..Nice approach sir – Akash Sarkar Feb 03 '19 at 13:43
  • @Peter One can discuss whether that's without induction. Sure, there is no explicit induction with a base case and an induction step, but something is happening inside those $\cdots$, and it is very induction-like. – Arthur Feb 03 '19 at 14:16
  • I agree in the case of $1+2^3+\cdots +n^3$ (I would not have an idea to show it without induction) , but the above sum-method does not use induction at all. We have $n$ columns with sum $n+1$, where is induction involved in this approach ? – Peter Feb 03 '19 at 14:19
  • @Peter There are many places induction can hide in an argument like that. For one thing, addition, multiplication and the natural numbers themselves are technically defined using induction so you can't use them without inducing somewhere. But even without that, you have things like showing that all the columns indeed do sum to $n+1$, that there are $n$ of them and that $(n+1) + (n+1) + \cdots + (n+1) = n(n+1)$ all look suspiciously like they have induction hidden deep, deep down. "Without induction" is a really hairy claim if you get pedantic. – Arthur Feb 03 '19 at 14:29
  • @Arthur I agree that it is pedantic. I do not think that the intent of the question is to apply a formal proof that addition and multiplication works (and to show THIS with induction). And if we accept that they do, no induction whatsoever remains. But since we need induction for the other part, this discussion is meaningless anyway. – Peter Feb 03 '19 at 14:34

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Hint: Expand the right side as $((1+2+3+\ldots+n)+(n+1))^2$ The square of the first term is the sum of the cubes up to $n$ by the induction hypothesis. The cross term is $2(n+1)\frac 12n(n+1)$

Ross Millikan
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