Summation Notation — Question 2

Question 2

Prove that for any positive integer $n$, $$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.$$

Question 2 - Solution

We prove the formula using mathematical induction.

Base Case:

For $n=1$, $$\sum_{k=1}^{1} k^2 = 1^2 = 1,$$ and $$\frac{1(1+1)(2\cdot1+1)}{6}=\frac{6}{6}=1.$$ Thus, the formula holds for $n=1$.

Inductive Hypothesis:

Assume that for some positive integer $n$, $$\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}.$$

Inductive Step:

Consider the sum up to $n+1$: $$\sum_{k=1}^{n+1} k^2 = \sum_{k=1}^{n} k^2 + (n+1)^2.$$

Using the inductive hypothesis, $$\sum_{k=1}^{n+1} k^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2.$$

Factor out $(n+1)$: $$\sum_{k=1}^{n+1} k^2 = (n+1)\left(\frac{n(2n+1)}{6} + (n+1)\right).$$

Combine terms inside the parentheses: $$\frac{n(2n+1)+6(n+1)}{6} = \frac{2n^2+7n+6}{6} = \frac{(2n+3)(n+2)}{6}.$$

Thus, $$\sum_{k=1}^{n+1} k^2 = \frac{(n+1)(n+2)(2n+3)}{6}.$$

This matches the given formula with $n$ replaced by $n+1$.

By mathematical induction, the formula holds for all positive integers $n$.

$$\boxed{\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}}$$