You are trying to prove that $\displaystyle \sum_{i=1}^n (2i- 1)= (n- 1)^3+ n^3$?
Your fundamental problem is that this isn't true!
When n= 2, $\displaystyle \sum_{i= 1}^2 (2i- 1)= (2(1)- 1)+ (2(2)-1)= 1+ 3= 4$ while $\displaystyle (2- 1)^3+ 2^3= 1+ 8= 9$.
Hi,
"Plug in" n = 1. This then reads 1 = 1, good so far. Next try n = 2: 1 + 3 = 1^{3} + 2^{3}, which is of course false. So your formula is not true for all n and so certainly can't be proved by induction.
$\displaystyle \sum_{i=1}^n2i-1=n^2$
The above is the correct formula and can be proved by induction. I suggest you try it.
Edit:
It suddenly occurred to me that you probably copied the problem incorrectly. What is true:
$\displaystyle \sum_{i=1}^n(2i-1)^2=(n-1)^3+n^3$
This formula is amenable to induction. Try it.
Hello, jpab29!
There must be typo . . .
Can this be proven by induction? . No!
. . $\displaystyle \displaystyle \sum^n_{i=1}(2i-1) \;=\;(n-1)^3 + n^3$ . This is not true!
$\displaystyle \text{Fact: }\;\sum^n_{i=1}(2i-1) \;=\;1 + 3 + 5 + \cdots + (2n-1) \;=\;{\color{blue}n^2}$
oh that's why. haha Actually it is derived from a conjecture that I am working on.
It is from this link.
Mathematical Induction
This is what I noticed:
For each line 1) - 4) (and letting the line number be $\displaystyle n$), I noticed that the first number in the sum is:
$\displaystyle (n-1)^2+1=n^2-2n+2$
Now, we then are adding the next $\displaystyle 2n-2=2(n-1)$ consecutive integers. As you observed, the sum is shown to be equal to $\displaystyle (n-1)^3+n^3$. So, we may state this as:
$\displaystyle \sum_{k=0}^{2(n-1)}(n^2-2n+2+k)=(n-1)^3+n^3$
Does this make sense?
Using the conjecture I stated originally (I had a typo the second time...):
$\displaystyle \sum_{k=0}^{2(n-1)}(n^2-2n+2+k)=(n-1)^3+n^3$
We have for:
$\displaystyle n=1$
$\displaystyle \sum_{k=0}^{2(1-1)}(1^2-2(1)+2+k)=(1-1)^3+1^3$
$\displaystyle 1=0^3+1^3$
$\displaystyle n=2$
$\displaystyle \sum_{k=0}^{2(2-1)}(2^2-2(2)+2+k)=(2-1)^3+2^3$
$\displaystyle 2+3+4=1^3+2^3$
$\displaystyle n=3$
$\displaystyle \sum_{k=0}^{2(3-1)}(3^2-2(3)+2+k)=(3-1)^3+3^3$
$\displaystyle 5+6+7+8+9=2^3+3^3$
$\displaystyle n=4$
$\displaystyle \sum_{k=0}^{2(4-1)}(4^2-2(4)+2+k)=(4-1)^3+4^3$
$\displaystyle 10+11+12+13+14+15+16=3^3+4^3$
If I was going to prove this by induction, this is what I would do. We have already shown the base case $\displaystyle P_1$ is true, so the induction hypothesis $\displaystyle P_n$ is the so-called conjecture:
$\displaystyle \sum_{k=0}^{2(n-1)}(n^2-2n+2+k)=(n-1)^3+n^3$
Let's simplify a bit:
$\displaystyle \sum_{k=0}^{2(n-1)}(n^2-2n+2)+\sum_{k=0}^{2(n-1)}(k)=(n-1)^3+n^3$
$\displaystyle (2n-1)(n^2-2n+2)+\sum_{k=0}^{2(n-1)}(k)=(n-1)^3+n^3$
$\displaystyle 2n^3-5n^2+6n-2+\sum_{k=0}^{2(n-1)}(k)=2n^3-3n^2+3n-1$
$\displaystyle \sum_{k=0}^{2(n-1)}(k)=2n^2-3n+1$
As the inductive step, I would add $\displaystyle 2n-1+2n=4n-1$:
$\displaystyle \sum_{k=0}^{2((n+1)-1)}(k)=2n^2+n=2(n+1)^2-3(n+1)+1$
We have derived $\displaystyle P_{n+1}$ from $\displaystyle P_n$ thereby completing the proof by induction.