Summation Notation — Question 3

Question 3 - Solution

We prove the formula using mathematical induction.

Base Case:

For $n=1$, $$\sum_{k=1}^{1} k^3=1^3=1, \qquad \left(\frac{1(1+1)}{2}\right)^2=1.$$ Thus, the formula holds for $n=1$.

Inductive Hypothesis:

Assume that for some positive integer $n$, $$\sum_{k=1}^{n} k^3=\left(\frac{n(n+1)}{2}\right)^2.$$

Inductive Step:

Consider the sum up to $n+1$: $$\sum_{k=1}^{n+1} k^3 = \sum_{k=1}^{n} k^3+(n+1)^3.$$

Using the inductive hypothesis, $$\sum_{k=1}^{n+1} k^3 = \left(\frac{n(n+1)}{2}\right)^2+(n+1)^3.$$

Factor out $(n+1)^2$: $$\sum_{k=1}^{n+1} k^3 = (n+1)^2\left(\frac{n^2}{4}+n+1\right).$$

Combine terms: $$\frac{n^2+4n+4}{4} = \frac{(n+2)^2}{4}.$$

Thus, $$\sum_{k=1}^{n+1} k^3 = (n+1)^2\cdot\frac{(n+2)^2}{4} = \left(\frac{(n+1)(n+2)}{2}\right)^2.$$

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

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

$$\boxed{\sum_{k=1}^{n} k^3=\left(\frac{n(n+1)}{2}\right)^2}$$