Prove: Positive Integer n Sum Equation

[markosheehan]

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.

 

[MarkFL]

Okay, so our induction hypothesis $P_n$ is:

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

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

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

\(\displaystyle 4(1)-3=2-1\)

\(\displaystyle 1=1\quad\checkmark\)

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

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

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

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

So, what we need to do is demonstrate:

\(\displaystyle n(2n-1)+4(n+1)-3=(n+1)(2(n+1)-1)\)

Can you proceed?

 

[markosheehan]

thanks

 

[MarkFL]

\(\displaystyle n(2n-1)+4(n+1)-3=2n^2+3n+1=(2n+1)(n+1)=(n+1)(2(n+1)-1)\)

Thus, we may state:

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

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