Proving $1+3+\cdots+(2n-1) = n^2$ using induction

Question

$ 1+3+\cdots+(2n-1) = n^2$

Base case; $n=1$, holds true since $1=1$ Induction Hypothesis; Suppose $ 1+3+\cdots+(2k-1) = k^2$

Then;

I have trouble with $2k+1$ do i add $2k+1$ to both sides or multiply them?

Answer 1

Hint: One has $$(k+1)^2-k^2=2k+1.$$

Answer 2

You need to show that $1+3+\cdots+(2k+1)=(k+1)^2$

Answer 3

You don't really need to use recursion. Just note that $1+3+\cdots+(2n-1)=(1+\cdots+2n)-2(1+\cdots+n)$.

The result follows from the well-known formula: for every integer $k$, $(1+\cdots+k)=k(k+1)/2$.

Answer 4

Let $k\ge 1$ such that

$$S (k)=1+3+\cdots+ (2k-1)=k^2$$ then

$$S (k+1)=1+3+\cdots+ (2(k+1)-1)$$

$$=1+3+\cdots+ (2k-1)+(2k+1) $$ $$=S (k)+(2k+1) $$ $$=k^2+2k+1=(k+1)^2$$

thus

$$\forall k\ge 1 \;\; \;S (k)=k^2$$

Answer 5

$$\begin{align}1+3+5+\cdots+(2k-1)+(2k+1)&=(1+3+5+\cdots+(2k-1))+(2k+1)\\&=k^2+(2k+1)\\&=k^2+2k+1\\&=(k+1)^2\end{align}$$ Note, $1+3+5+\cdots+(2k-1)=k^2$ is by the induction step.

Answer 6

$$1+\cdots+(2k-1)+(2k+1)=k^2+(2k+1)=(k+1)^2$$