Proof by Mathematical Induction

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The discussion focuses on proving the formula for the sum of the first n odd numbers using mathematical induction. The base case is established as true with S_1 equaling 1, confirming that 1^2 equals 1. The induction hypothesis is stated as S_k equaling k^2. The inductive step involves adding 2k+1 to both sides, leading to the expression S_k + 2k + 1, which simplifies to (k+1)^2. This successfully demonstrates that if the hypothesis holds for k, it also holds for k+1, completing the proof.
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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
 
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We are given to prove:

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

Step 1: demonstrate the base case is true.

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

This is true.

Step 2: State the induction hypothesis $P_k$:

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

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

What do you get if you add $2k+1$ to both sides?
 
MarkFL said:
We are given to prove:

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

Step 1: demonstrate the base case is true.

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

This is true.

Step 2: State the induction hypothesis $P_k$:

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

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 $P_{k+1}$.
 
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:

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

Can you incorporate the added terms into the sum on the left and factor the right?
 
Since several days has gone by with no feedback from the OP, I will finish. We have:

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

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

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

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

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

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

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