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Conjecture formula from following equations, and prove conjecture: $1=1,\\2+3+4=1+8,\\5+6+7+8+9=8+27,\\10+11+12+13+14+15+16=27+64\\$

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

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

Need to show $S(n+1)=\sum_{i=n^2+1}^{(n+1)^2}i=n^3+(n+1)^3$? Not sure. Thanks.

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2 Answers 2

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Notice that (we can prove it using Gauss method) $$\sum_{k=p}^q k=\frac{(q-p+1)(p+q)}2$$ so we find that

$$S(n)=\frac{(n^2-(n-1)^2)(n^2+(n-1)^2+1)}{2}=2\,{n}^{3}−3\,{n}^{2}+3\,n−1\\=n^3+(n-1)^3$$

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  • $\begingroup$ No need for induction? I think I am supposed to prove it using induction. I guess not if Gauss is already proven using induction? $\endgroup$
    – miniparser
    Sep 16, 2014 at 19:17
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How about \begin{align} S(n) &=\sum_{i=1}^{n^2}i - \sum_{i=1}^{(n-1)^2}i\\ &=\dfrac{n^2(n^2+1)}2 - \dfrac{[(n-1)^2]\,[(n-1)^2 + 1]}2 \\ &=\dfrac{n^4+n^2}2 - \dfrac{(n-1)^4 + (n-1)^2}2 \\ &=\dfrac{n^4-(n-1)^4+n^2-(n-1)^2}2 \\ &=\dfrac{(n^2-(n-1)^2)(n^2+(n-1)^2)+(n-(n-1))(n+(n-1))}2 \\ &=\dfrac{(2n-1)(2n^2-2n+1)+(2n-1)}2 \\ &=\dfrac{(2n-1)(2n^2-2n+2)}2 \\ &=(2n-1)(n^2-n+1) \\ &=2n^3 -3n^2 + 3n - 1 \\ &=n^3 + (n^3 - 3n^2 + 3n - 1) \\ &=n^3 + (n-1)^3 \\ \end{align}

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