How to expand summations using mathematical induction

Question

By Induction $$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$ But I didn't understand how I can do this, for this function or another function, i.e how to derive the right hand side function for the summation.

Answer 1

There is a certain amount of guessing involved. You start by looking at similar formulas and looking for a pattern.

\begin{align} \sum_{i=0}^n i^0 = n+1 \\ \sum_{i=0}^n i^1 = \dfrac 12 n^2 + \dfrac 12 n \\ \end{align}

It's not a lot, but it suggests the hypothesis

$$\sum_{i=0}^n i^2 = An^3 + Bn^2 + Cn+D \tag{1}$$

for some numbers $A,B,C$, and $D$.

The first step is to make equation $(1)$ true for $n=0$.

\begin{align} \sum_{i=0}^0 i^2 &= A0^3 + B0^2 + C0+D \\ 0 &= D \\ D &= 0 \end{align}

So we can now replace equation $(1)$ with

$$\sum_{i=0}^n i^2 = An^3 + Bn^2 + Cn \tag{2}$$

The next induction step is to assume that equation $(2)$ is true for some particular value, say $n = k$ and then make sure that it also true for $n = k+1$.

\begin{align} \sum_{i=0}^{k+1} i^2 &= \sum_{i=0}^{k} i^2 + (k+1)^2 \\ A(k+1)^3 + B(k+1)^2 + C(k+1) &= Ak^3 + Bk^2 + Ck + (k^2+2k+1) \\ Ak^3 +(3A+B)k^2 + (3A+2B+C)k + (A+B+C) &= Ak^3 + (B+1)k^2 + (C+2)k \\ (3A-1)k^2 + (3A+2B-2)k + (A+B+C-1) &= 0 \\ \hline 3A-1 &= 0 \\ 3A+2B-2 &= 0 \\ A+B+C-1 &= 0 \\ \hline A &= \dfrac 13 \\ B &= \dfrac 12 \\ C &= \dfrac 16 \end{align}

Hence

\begin{align} \sum_{i=0}^n i^2 &= \dfrac 13 n^3 + \dfrac 12 n^2 + \dfrac 16 n \\ &= \dfrac 16(2n^3 + 3n^2 + n) \\ &= \dfrac 16 n(2n^2 + 3n + 1) \\ &= \dfrac 16 n(n+1)(2n+1) \\ \end{align}