Find and prove formula

Question

For $n\in \Bbb N$, find and prove a formula for $$\sum_{i=1}^n(2i-1)$$

How do I use induction to find a formula for it? I haven't not encountered any question like this in my textbook.

Answer 1

HINT: If you have not yet formed a conjecture about what it might be, here are the first couple terms: $$\sum_{i=1}^1 (2i-1)=1$$ $$\sum_{i=1}^2 (2i-1)=4$$ $$\sum_{i=1}^3 (2i-1)=9$$ $$\sum_{i=1}^4 (2i-1)=16$$

Do you see a pattern? Once you have formed a conjecture, I can give you some tips about how to prove your conjecture using induction.

Answer 2

Use $1+2+3+...+n=\frac{n(n+1)}{2}$. Then $$\sum_{i=1}^{n}(2i-1)=2\sum_{i=1}^{n}i-n=2\frac{n(n+1)}{2}-n=n^2+n-n=n^2$$

Answer 3

In order to go by induction, you first need a formula.

Start with $n=1$, then the sum is $1$.

For $n=2$, the sum is $1+3 = 4$.

For $n=3$, the sum is $1+3+5 = 9$.

With this in mind, you can see that the sum "seems" to be $n^2$ (it's an intelligent guess , it's difficult to say how you would do this in a more general situation). This can be proved inductively.

Base case : We know that for $n=1$, the sum is $n^2 = 1^2 = 1$.

Inductive Case : Suppose that this is true for $n$ i.e. $\sum_{i=1}^n (2i-1) = n^2$. Simply add $2n+1$ to both sides, and note that $2n+1 = 2(n+1) - 1$. $$ \sum_{i=1}^{n+1} (2i-1) = \sum_{i=1}^n (2i-1) + (2(n+1)-1) = n^2 + 2n+1 = (n+1)^2 $$

Hence, the proposition is true for $n+1$, hence the proposition is true for all $n$ by induction.

The intuition for guessing $n^2$ comes from the well known formula that $\sum_{i=1}^n i = \frac{n(n+1)}{2}$. Substituting this formula in the expression $\sum (2i-1)$ gave me back $n^2$, so I could use this for induction.