Question

Prove that 1 + 3 + 5 + … + (2n – 1) = n2 using the principle of Mathematical induction.

Answer 1

Need-to-know

• Mathematical Induction(Proofs)

It is used to show a claim. It is used when proof over natural numbers is enough.

The proof is not perfect, though. It is likened to the dominos.

Proof

Claim: $\bold{1 + 3 + 5 + ... + (2n - 1) = n^2}$

Let's check if $n=1$ works.

$\bold{(n=1)}$ $1=1^2$

Now, let's see if $k=n+1$ works, the reason is $k$ is the next integer of $n$. Only one value that works is enough, because

$\bold{(k=n+1)}$ $1+3+5+...+(2n+1)=(n+1)^2$

This proves every next integer of $1$ works. Hence, it is proven for all the natural numbers.

More information

• Series

$\begin{cases} & S =1+3+5+...+(2n-1) \\ & S =(2n-1)+...+5+3+1 \end{cases}$

$\Longleftrightarrow 2S=n(2n)$

$\therefore S=n^2$ is the series.

• Facts

The mathematical induction is not actually inductive. It is deductive because the conclusion comes from normal facts.

If it were inductive, we would have proven for every natural number.

Answer 2

Given :

$\sf 1+3+5+....+(2n-1) = n^2$

To Find :

• Prove the given equation using Principal of mathematical induction

Method :

Let $\sf P(n)$ be a given statement for $n \in N.$

Initial Step : Let the statement is true for $\sf n=1\:i.e.,P(1)$ is true and

Inductive Step : If the statement is true for $\sf n = k$ (where k is a particular but arbitrary natural number) then the statement is true for $\sf n=k+1.\: i.e.,$ truth of $\sf P(k)$ implies the truth of $\sf P(k+1)$. Then

Solution :

Let the given statement $\sf P(n)$ be defined as

$\sf 1+3+5+....+(2n-1) = n^2\:for\:n \in N$ Then$\sf P(n)$ is true for all natural numbers $\sf n$

Step 1 :- $\sf Put \: n=1$

$\longrightarrow\sf LHS \: P(1) =1\\\\\sf\longrightarrow RHS=1^2\Rightarrow 1\\\\\sf\longrightarrow LHS = RHS$

$\sf \therefore P(1)\:is\:true.$

Step 2 :- Let us assume that the statement is true for $\sf n=k$

$\sf i . e.,P(k) is true$

$\sf 1+3+5+... (2k-1)=k^2\:is\:true$

Step 3 :- To prove that $\sf p(k+1)$ is true.

$\sf P(k+1) = 1+3+5+...+(2k-1)+(2k+1)\\\\\rightarrow \sf p(k)+(2k+1)\\\\\rightarrow\sf k^2+2k+1\\\\\rightarrow \sf (k+1)^2$

$\sf P(k+1)$ is true whenever $\sf P(k)$ is true.

$\sf P(n)$ is true for all $\sf n \in N$