Exercise 4.3.10 ($\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$)

Question

By use of the algebraic identity $(x+1)^3 - x^3 = 3x^2 + 3x + 1$ establish that $(n+1)^3 - 1^3 = \sum_{x=1}^n {(x+1)^3 - x^3} = \sum_{x=1}^n (3x^2 +3x +1)$, and so derive the result $\sum_{x=1}^n x^2 = n(n+1)(2n+1)/6$.

Richard Ganaye · Accepted Answer

\begin{proof} 
With some ``copy and paste'' from Problem 9, we obtain
\begin{align*}
\sum_{i=1}^n [(i+1)^3 - i^3]&= \sum_{i=1}^n (i+1)^3  - \sum_{i=1}^n i^3 \
&=\sum_{j = 2}^{n+1} j^3 - \sum_{j=1}^n j^3&(j=i +1 	ext{ in the first sum})\
&=\left((n+1)^3 + \sum_{j = 2}^{n} j^3 ight) - \left( 1 + \sum_{j=1}^n j^3 ight)\
&=(n+1)^3 - 1.
\end{align*}
Since $(i+1)^3 - i^3 = 3i^2 + 3i + 1$ for all $i$, we obtain
\begin{align}
\sum_{i=1}^n (3i^2 + 3i + 1)= (n+1)^3 - 1.
\end{align}
We define $S(n) = \sum_{i=1}^n i^2$ for all positive integer $n$. By (1), and the result of Problem 9,
\begin{align*}
(n+1)^3 - 1 &= \sum_{i=1}^n (3i^2 + 3i + 1)\
&= 3 \sum_{i=1}^n i^2 + 3\sum_{i=1}^n i + \sum_{i=1}^n 1\
&=3 S(n) + 3 \frac{n(n+1)}{2} + n.
\end{align*}
Therefore
\begin{align*}
3\left[ S(n) + \frac{n(n+1)}{2}ight] = (n+1)^3 - (n+1) = n(n+1)(n+2),
\end{align*}
thus
\begin{align*}
S(n) &= \frac{n(n+1)(n+2)}{3} - \frac{n(n+1)}{2}\
&= \frac{n(n+1)(2n+1)}{6}.
\end{align*}
For all positive integer $n$,
$$\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$$

\end{proof}

Exercise 4.3.10 ($\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$)

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Exercise 4.3.10 ($\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.$)

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