I'm asked to show that Σ k^3 = [(n(n+1))/2]^2, ∀n ∈ ℕn
Can anyone help?k=1
Hi Lindsai,
P(k)
$\displaystyle \sum_{} k^3=\left(\frac{k(k+1)}{2}\right)^2$
P(k+1)
$\displaystyle \sum_{} (k+1)^3=\left(\frac{(k+1)(k+2)}{2}\right)^2$
Test if P(k) true causes P(k+1) true
Proof
If P(k) is true, then the sum of k+1 cubes is
$\displaystyle \sum_{} k^3+(k+1)^3=\left(\frac{k(k+1)}{2}\right)^2+(k+1)^ 3$
Now you only need verify that this is what you get when you expand P(k+1).
Then test the formula for the very first cube.
$\displaystyle \frac{(k+1)(k+1)}{2}\ \frac{(k+2)(k+2)}{2}=\frac{(k+2)(k+2)(k+1)^2}{2^2}$
$\displaystyle =\frac{\left(k^2+4k+4\right)(k+1)^2}{2^2}=\left(\f rac{k(k+1)}{2}\right)^2+4(k+1)\left(\frac{k+1}{2}\ right)^2$
$\displaystyle =\left(\frac{k(k+1)}{2}\right)^2+(k+1)^3$
test for n=1
$\displaystyle 1^3=\left(\frac{1(1+1)}{2}\right)^2$
What????
Does she have to do it the
1. test for the first term
2. test for n+1
way ?
that's just a format for anyone who wants an answer without knowing much about
induction.
As I do not advocate that, I do't do it that way.
You're welcome to of course!
The inductive step discovers a term-by-term relationship.
It finds out if the following relationship exists....
true for k=1 causes true for k=2
true for k=2 causes true for k=3
true for k=3 causes true for k=4
to infinity.
It's like stacking up an endless sequence of dominoes.
Then, if you topple the first domino, they all fall.
Testing for "true for the 1st term" at that stage then proves the formula is true to infinity.
To say you must prove for T1 first and then do the inductive step
and declaring that that's the way only presents a compressed dumbed-down version.