Finite sum $\sum_{j=0}^{n-1} j^2$

Question

How can I calculate this finite sum? Can someone help me?

$$\sum_{j=0}^{n-1} j^2$$

Answer 1

I like this explanation:

Each row in the first triangle sums to $j^2$
So the sum of the rows is the number we seek.

Take that triangle and rotate left and right. The sum is of the three trianges, then is 3 times the number we seek. But when we sum it up, we get $(2n+1)$ in every entry in the triange or $(2n+1)$ times the number of entries.

$3\sum_\limits{j = 1}^n j^2 = (2n+1)\sum_\limits{j = 1}^n j$

Answer 2

Notice $$\begin{align} j^2 &= \frac12[ j(j+1) + (j-1)j ]\\ &= \frac12\left[\frac{\color{red}{(j+2)}-\color{green}{(j-1)}}{3}j(j+1) + \frac{\color{blue}{(j+1)}-\color{magenta}{(j-2)}}{3}(j-1)j\right]\\ &= \frac16\left[\left( \underbrace{j(j+1)\color{red}{(j+2)}}_A - \underbrace{\color{green}{(j-1)}j(j+1)}_B \right) + \left( \underbrace{(j-1)j\color{blue}{(j+1)}}_C - \underbrace{\color{magenta}{(j-2)}(j-1)j}_D \right)\right]\\ &= \frac16\left[ \underbrace{j(j+1)(2j+1)}_{A+C} - \underbrace{(j-1)j(2j-1)}_{B+D}\right] \end{align} $$ The sum at hand is a telescoping sum with

$$\sum_{j=1}^{n-1}j^2 = \frac{(n-1)n(2n-1)}{6}$$

Answer 3

You can prove by induction that $\sum\limits_{j=0}^{n-1}j^2=(2n^3-3n^2+n)/6$.

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First, show that this is true for $n=1$:

$\sum\limits_{j=0}^{1-1}j^2=(2-3+2)/6$

Second, assume that this is true for $n$:

$\sum\limits_{j=0}^{n-1}j^2=(2n^3-3n^2+n)/6$

Third, prove that this is true for $n+1$:

$\sum\limits_{j=0}^{n}j^2=$

$\color\red{\sum\limits_{j=0}^{n-1}j^2}+n^2=$

$\color\red{(2n^3-3n^2+n)/6}+n^2=$

$[2(n+1)^3-3(n+1)^2+(n+1)]/6$

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Please note that the assumption is used only in the part marked red.

Answer 4

Try to break a cube into L-shells, then break each L-shell into $1$, $n$, and $n^2$ components.

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\left.\begin{array}{rclcl} \ds{j^{\underline{2}}} & \ds{=} & \ds{j\pars{j - 1}} & \ds{=} & \ds{j^{2} - j} \\[1mm] \ds{j^{\underline{1}}} & \ds{=} & \ds{j} \end{array}\right\} \implies j^{2} = j^{\underline{2}} + j^{\underline{1}}}$

\begin{align} \sum_{j = 0}^{n - 1}j^{2} & = \sum_{j = 0}^{n - 1}\pars{j^{\underline{2}} + j^{\underline{1}}} = {1 \over 3}\,n^{\underline{3}} + {1 \over 2}\,n^{\underline{2}} = {1 \over 3}\,n\pars{n - 1}\pars{n - 2} + {1 \over 2}\,n\pars{n - 1} \\[5mm] & = n\pars{n - 1}\pars{{n - 2 \over 3} + {1 \over 2}} = \bbx{\ds{n\pars{n - 1}\pars{2n - 1} \over 6}} \end{align}

Answer 6

This problem has many answers:

HINT : We define $a_{n-1}=$$\sum_{j=0}^{n-1} j^2$ So we need to solve the following recurrent sequence

$a_{n}=$ $a_{n-1}+n^2$