Proof by Mathematical Induction

[needmathhelp1]

I am confused on a math problem. I am supposed to use mathematical induction to show that

(Summation with n on top and i=1 on the bottom) (2i-1) = 1+3+5+...+(2n-1)= n^2

 

[MarkFL]

We are given to prove:

\(\displaystyle S_n=\sum_{i=1}^n(2i-1)=n^2\)

Step 1: demonstrate the base case is true.

\(\displaystyle S_1=\sum_{i=1}^1(2i-1)1=1^2\)

This is true.

Step 2: State the induction hypothesis $P_k$:

\(\displaystyle S_k=\sum_{i=1}^k(2i-1)=k^2\)

Step 3: Make the inductive step, and obtain $P_{k+1}$.

What do you get if you add $2k+1$ to both sides?

 

[needmathhelp1]

We are given to prove:

\(\displaystyle S_n=\sum_{i=1}^n(2i-1)=n^2\)

Step 1: demonstrate the base case is true.

\(\displaystyle S_1=\sum_{i=1}^1(2i-1)1=1^2\)

This is true.

Step 2: State the induction hypothesis $P_k$:

\(\displaystyle S_k=\sum_{i=1}^k(2i-1)=k^2\)

Step 3: Make the inductive step, and obtain $P_{k+1}$.

What do you get if you add $2k+1$ to both sides?

I understand the first two steps, but I am not sure what the inductive step is or how to obtain $P_{k+1}$.

 

[MarkFL]

The inductive step is essentially a legal algebraic operation you perform on $P_k$ that leads to $P_{k+1}$, usually after some simplification.

If you use the step I suggested, you then have:

\(\displaystyle \sum_{i=1}^k(2i-1)+2k+1=k^2+2k+1\)

Can you incorporate the added terms into the sum on the left and factor the right?

 

[MarkFL]

Since several days has gone by with no feedback from the OP, I will finish. We have:

\(\displaystyle \sum_{i=1}^k(2i-1)+2k+1=k^2+2k+1\)

On the left side, rewrite the added terms so that they are in the form of the summand:

\(\displaystyle \sum_{i=1}^k(2i-1)+2(k+1)-1=k^2+2k+1\)

Incorporate these terms into the sum and factor the right side as a square:

\(\displaystyle \sum_{i=1}^{k+1}(2i-1)=(k+1)^2\)

We have now derived $P_{k+1}$ from $P_k$, thereby completing the proof by induction.