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Solution

(related to Problem: Sum of Consecutive Squares)

• Note that we can write $x^2$ using the falling factorial powers as $x^2=x(x-1)+x=x^{\underline{2}}+x^{\underline{1}}.$
• Now, $$\begin{align}1^2+2^2+3^2+\ldots+n^2&=\sum_{x=1}^n (x^{\underline{2}}+x^{\underline{1}})\nonumber\\ &=\left(\frac{x^{\underline 3}}{3}+\frac{x^{\underline 2}}{2}\right)\;\Rule{1px}{4ex}{2ex}^{n+1}_{1}\nonumber\\ &=\left(\frac{x(x-1)(x-2)}{3}+\frac{x(x-1)}{2}\right)\;\Rule{1px}{4ex}{2ex}^{n+1}_{1}\nonumber\\ &=\frac{(n+1)n(n-1)}{3}+\frac{(n+1)n}{2}\nonumber\\ &=\frac{n(n+1)(2n+1)}{6}.\nonumber \end{align}$$

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References

Bibliography

1. Miller, Kenneth S.: "An Introduction to the Calculus of Finite Differences And Difference Equations", Dover Publications, Inc, 1960