MHB Prove: Positive Integer n Sum Equation

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The discussion focuses on proving the equation for the sum of a series of positive integers: 1 + 5 + 9 + 13 + ... + (4n - 3) = n/2(4n - 2) using mathematical induction. The base case for n=1 is verified successfully, showing that the equation holds true. The induction hypothesis is established, and the next step involves demonstrating that if the equation holds for n, it also holds for n+1. The participants successfully derive the expression for n+1, confirming the induction step. The proof is completed, establishing the validity of the equation for all positive integers n.
markosheehan
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prove by induction for all positive integers n: 1+5+9+13+...+(4n-3)= n/2(4n-2)
i tried this by trying to prove n/2(4n-2)+ (4(k+1)-3) = k+1/2(4(k+1)-2) but it did not work out for me.
 
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Okay, so our induction hypothesis $P_n$ is:

$$\sum_{k=1}^{n}(4k-3)=\frac{n}{2}(4n-2)=n(2n-1)$$

First we should verify that the base case $P_1$ is true:

$$\sum_{k=1}^{1}(4n-3)=(1)(2(1)-1)$$

$$4(1)-3=2-1$$

$$1=1\quad\checkmark$$

Okay, the base case is true. So, for our induction step, let's write:

$$\sum_{k=1}^{n}(4k-3)+4(n+1)-3=n(2n-1)+4(n+1)-3$$

Incorporating the added term into the sum on the left, we have:

$$\sum_{k=1}^{n+1}(4k-3)=n(2n-1)+4(n+1)-3$$

So, what we need to do is demonstrate:

$$n(2n-1)+4(n+1)-3=(n+1)(2(n+1)-1)$$

Can you proceed?
 
thanks
 
$$n(2n-1)+4(n+1)-3=2n^2+3n+1=(2n+1)(n+1)=(n+1)(2(n+1)-1)$$

Thus, we may state:

$$\sum_{k=1}^{n+1}(4k-3)=(n+1)(2(n+1)-1)$$

This is $P_{n+1}$, which we have derived from $P_n$, thereby completing the proof by induction. :D
 
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