Summation Notation — Question 1

Question 1 - Solution

We prove the formula using mathematical induction.

Base Case:

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

Inductive Hypothesis:

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

Inductive Step:

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

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

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

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

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

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

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