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if I have 4 equations..

$$ 1=1$$ $$2+3+4=1+8$$ $$5+6+7+8+9=8+27$$ $$10+11+12+13+14+15+16=27+64$$

how do I find the general formula (that is suggested by the equations) using informal inductive reasoning?

how do I prove that the closed formula is correct using mathematical induction?

thanks

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2 Answers

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Hint:

I propose this formula: $$\sum_{k=n^2+1}^{(n+1)^2}k=n^3+(n+1)^3.$$

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The fist equation is misleading, it should be $1=0+1$ to fit with the general pattern.

This said, you observe on the LHS that there are $1,3,5,7\cdots=2n-1$ terms, and they form the sequence of naturals. On the RHS, you have the sum of two consecutive cubes,$(n-1)^3+n^3$.

Using the formula for the sum of odd numbers, the equations can be written

$$S_n:=\sum_{i=(n-1)^2+1}^{n^2}i=(n-1)^3+n^3.$$

The proof by induction isn't straightforward, because when you go from $n$ to $n+1$, all $2n-1$ terms in the sum are increased by $2n-1$, and two more terms are added.

Hence,

$$S_{n+1}=S_n+(2n-1)(2n-1)+(n+1)^2-1+(n+1)^2=S_n+\color{blue}{6n^2+2}.$$

On the other hand,

$$n^3+(n+1)^3=(n-1)^3+n^3+\color{blue}{6n^2+2}.$$

Any of the given equations but the first can be taken for the base case.

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More formally,

$$S_{n+1}=\sum_{i=n^2+1}^{(n+1)^2}i=\sum_{i=(n-1)^2+2n-1+1}^{n^2+2n-1+2}i=\sum_{i=(n-1)^2+1}^{n^2}(i+2n-1)+(n+1)^2-1+(n+1)^2.$$

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