ankarskogC

2021-08-18

Prove the following statement using mathematical induction or disapprove by counterexample.
If you use mathematical induction, then you should explain each step and you should highlight the inductive hypothesis, etc. Explaining each step is very important.
$1+5+9+13+\cdots +\left(4n-3\right)=\frac{n\left(4n-2\right)}{2}$

Alannej

Step 1
We need to prove that
$P\left(n\right)=1+5+9+13+\cdots +\left(4n-3\right)=\frac{\left(n\left(4n-2\right)\right)}{2}$
We use induction hypothesis to prove it.
For that we follow the following steps.
Let $P\left(n\right)$ be the given sum.
So,
(I) We prowe that $P\left(1\right)$ holds.
(II) We assume that it is true for $n=k.$ That means, let $P\left(k\right)$ be true.
(III) We use the hypothesis in the second statement to prove $P\left(k+1\right)$ is also true. Which futher implies, it is true for any value of n.
Step 2
Now,
Let $n=1,$ we have
$P\left(n\right)=1+5+9+13+\cdots +\left(4n-3\right)=\frac{\left(n\left(4n-2\right)\right)}{2}$
We have,
$1+5+9+13+\cdots +\left(4n-3\right)=\frac{\left(n\left(4n-2\right)\right)}{2}$
Left hand side $=1$
Right hand side $=\frac{\left(n\left(4n-2\right)\right)}{2}=\frac{\left(1\left(4\left(1\right)-2\right)\right)}{2}=1$
Hence, Left hand side $=$ Right hand side.
Therefore, it is true for $n=1$
Now, we assume that $P\left(n\right)$ is true for $n=k$, that is, $P\left(k\right)$ holds.
1) $1+5+9+13+\cdots +\left(4k-3\right)=\frac{\left(k\left(4k-2\right)\right)}{2}$
Let this be true.
Step 3
Now, let $n=k+1$
We need to prove the following equality for $n=k+1$
$1+5+9+13+\cdots +\left(4\left(k+1\right)-3\right)=\frac{\left(\left(k+1\right)\left(4\left(k+1\right)-2\right)\right)}{2}$
Left hand side:
$1+5+9+13+\cdots +\left(4\left(k+1\right)-3\right)=1+5+9+13+\cdots +\left(4k-3\right)+\left(4\left(k+1\right)-3\right)$
$=\left[1+5+9+13+\cdots +\left(4k-3\right)1+4\left(k+1\right)-3$
$=\frac{k\left(4k-2\right)}{2}+4\left(k+1\right)-3$
$=\frac{k\left(4k-2\right)}{2}+4k+1$

Do you have a similar question?