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$$\begin{align*} 1 &\leadsto 1 \\ 1+3 &\leadsto 2^2 \\ 1+3+5 &\leadsto 3^2 \end{align*}$$ In general, if $f(x)=2x+1$, then $f(0)+f(1)+f(2)...f(n)=(n+1)^2$.

Now, $$\begin{align*} 1 &\leadsto 1 \\ 1+7+19+\cdots &\leadsto m^3 \\ \end{align*}$$ Is there a function that generates these numbers?

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Just take the difference between the next, and the current number. For example $(x+1)^2-x^2=2x+1$. Likewise, $(x+1)^3-x^3=3x^2+3x+1$, which is the expression you're looking for.

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  • $\begingroup$ Exactly what I was looking for thanks! $\endgroup$ – AAron Jul 4 '15 at 7:01
  • $\begingroup$ But if the function, more vaguely, was x^n, then would the sum be (x+1)^n-x^n $\endgroup$ – AAron Jul 4 '15 at 7:02
  • $\begingroup$ @AAron That would work for any function, it is pretty intuitive that $\sum_{i=0}^n\left(f(i+1)-f(i)\right)=f(n+1)-f(0)$ $\endgroup$ – mniip Jul 4 '15 at 7:04

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