For all $n \ge 1$, prove that $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$.

Question

Accepted Answer

Let $P(n)$ be the statement
$$1^2+2^2+3^2+\ldots+n^2=\frac{n(n+1)(2n+1)}{6}.$$

**Base case:** For $n=1$,
$$	ext{LHS}=1^2=1,\qquad 	ext{RHS}=\frac{1(1+1)(2\cdot1+1)}{6}=\frac{1\cdot2\cdot3}{6}=1.$$
Since LHS $=$ RHS, $P(1)$ is true.

**Inductive step:** Assume $P(k)$ is true for some $k\ge1$, i.e.
$$1^2+2^2+\ldots+k^2=\frac{k(k+1)(2k+1)}{6}. \qquad 	ext{(Induction Hypothesis)}$$

We must show $P(k+1)$ is true, i.e.
$$1^2+2^2+\ldots+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}.$$

Starting from the LHS and using the induction hypothesis:
$$1^2+2^2+\ldots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$
$$=(k+1)\left[\frac{k(2k+1)}{6}+(k+1)ight]=(k+1)\cdot\frac{k(2k+1)+6(k+1)}{6}$$

Expand the bracket:
$$k(2k+1)+6(k+1)=2k^2+k+6k+6=2k^2+7k+6=(k+2)(2k+3)$$

So
$$1^2+2^2+\ldots+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{(k+1)\big((k+1)+1\big)\big(2(k+1)+1\big)}{6},$$
which is exactly $P(k+1)$.

Thus $P(k)\Rightarrow P(k+1)$.

> [!ANSWER] Since $P(1)$ is true and $P(k)\Rightarrow P(k+1)$ for every $k\ge1$, by the Principle of Mathematical Induction $P(n)$ is true for all $n\ge1$: $1^2+2^2+\ldots+n^2=\dfrac{n(n+1)(2n+1)}{6}$.

Q.For all $n \ge 1$ , prove that $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ .

Concept understanding — Proving Summation Formulas by Induction

Q.For all n≥1n \ge 1n≥1, prove that 12+22+32+42+…+n2=n(n+1)(2n+1)61^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = \dfrac{n(n+1)(2n+1)}{6}12+22+32+42+…+n2=6n(n+1)(2n+1)​.

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Q.For all $n \ge 1$ , prove that $1^2 + 2^2 + 3^2 + 4^2 + \ldots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$ .

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