Summation Notation — Question 4

Question 4 - Solution

We prove the identity using mathematical induction.

Base Case:

For $n=1$, $$\sum_{k=1}^{1} (2k-1) = 2(1)-1 = 1,$$ and $$1^2 = 1.$$ Thus, the statement holds for $n=1$.

Inductive Hypothesis:

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

Inductive Step:

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

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

Simplify: $$n^2 + 2n + 1 = (n+1)^2.$$

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

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

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

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