Tom's question at Yahoo Answers regarding proof by induction

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The discussion centers on proving the formula for the sum of the first n odd numbers using mathematical induction. The correct formula to prove is $$\sum_{i=1}^n(2i+1)=n^2+2n$$. The proof involves demonstrating the base case, establishing the induction hypothesis, and deriving the case for n+1. Mark provides a clear step-by-step explanation, confirming the validity of the proof by induction method.

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Here is the question:

Proove the following by mathematical induction?

while justifying your supporting arguments using the language of proof coherently, concisely and logically.
3 + 7 + 11 + 15 + ... to n terms = 2n^2 +n

Here is a link to the question:

Proove the following by mathematical induction? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Tom,

The statement as given is not true, so I assume it is a typo, and instead the problem should be as follows.

We are given to prove by induction the following:

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

Step 1: demonstrate the base case $P_1$ is true:

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

$$2(1)+1=1+2(1)$$

This is true.

Step 2: state the induction hypothesis $P_k$:

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

Step 3: derive $P_{k+1}$ from $P_k$ to complete the proof by induction.

Let our inductive step be to add $$2(k+1)+1$$ to both sides of $P_k$:

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

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

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

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

To Tom or any other guests viewing this topic, I invite and encourage you to register and post other proof by induction problems either in our http://www.mathhelpboards.com/f21/ or http://www.mathhelpboards.com/f15/ forums depending on the nature of the problem, or course from which it is given.

Best Regards,

Mark.
 

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