
#1
September 19th, 2016,
14:01
prove by induction for all positive integers n: 1+5+9+13+........+(4n3)= n/2(4n2)
i tried this by trying to prove n/2(4n2)+ (4(k+1)3) = k+1/2(4(k+1)2) but it did not work out for me.

September 19th, 2016 14:01
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#2
September 19th, 2016,
14:12
Okay, so our induction hypothesis $P_n$ is:
$ \displaystyle \sum_{k=1}^{n}(4k3)=\frac{n}{2}(4n2)=n(2n1)$
First we should verify that the base case $P_1$ is true:
$ \displaystyle \sum_{k=1}^{1}(4n3)=(1)(2(1)1)$
$ \displaystyle 4(1)3=21$
$ \displaystyle 1=1\quad\checkmark$
Okay, the base case is true. So, for our induction step, let's write:
$ \displaystyle \sum_{k=1}^{n}(4k3)+4(n+1)3=n(2n1)+4(n+1)3$
Incorporating the added term into the sum on the left, we have:
$ \displaystyle \sum_{k=1}^{n+1}(4k3)=n(2n1)+4(n+1)3$
So, what we need to do is demonstrate:
$ \displaystyle n(2n1)+4(n+1)3=(n+1)(2(n+1)1)$
Can you proceed?

#3
September 19th, 2016,
15:46
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Pessimist Singularitarian
#4
September 19th, 2016,
16:08
$ \displaystyle n(2n1)+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}(4k3)=(n+1)(2(n+1)1)$
This is $P_{n+1}$, which we have derived from $P_n$, thereby completing the proof by induction.