Can anyone help me prove (k(k-1)/2)^2 by mathematical induction
Hello, hats_06!
Please state the original equation.Can anyone help me prove (k(k-1)/2)^2 by mathematical induction?
What you're asking is: "Can anyone prove $\displaystyle 36$?"
I recognize the expression.
It may have come from: .$\displaystyle 1^3 + 2^3 + 3^3 + \cdots + (k-1)^3 \;=\;\left[\frac{k(k-1)}{2}\right]^2 $
You wish to prove
$\displaystyle \sum_{i = 1}^{k - 1}i^3 = \frac{k^2(k - 1)^2}{4}$
The base case is k = 2:
$\displaystyle \sum_{i = 1}^{1}i^3 = \frac{2^2(2 - 1)^2}{4} = 1$ (check!)
Now assume theorem is true for some value of k, say n. So we are assuming
$\displaystyle \sum_{i = 1}^{n - 1}i^3 = \frac{n^2(n - 1)^2}{4}$
Now we need to show that
$\displaystyle \sum_{i = 1}^{(n + 1) - 1}i^3 = \frac{(n + 1)^2((n + 1) - 1)^2}{4}$
or
$\displaystyle \sum_{i = 1}^{n}i^3 = \frac{(n + 1)^2n^2}{4}$
The LHS is:
$\displaystyle \sum_{i = 1}^{n}i^3 = \sum_{i = 1}^{n - 1} + n^3$
$\displaystyle = \frac{n^2(n - 1)^2}{4} + n^3$
by assumption.
$\displaystyle = \frac{n^2(n - 1)^2}{4} + \frac{4n^3}{4}$
$\displaystyle = \frac{n^2(n - 1)^2 + 4n^3}{4}$
$\displaystyle = \frac{n^2(n^2 - 2n + 1) + 4n^3}{4}$
$\displaystyle = \frac{n^4 - 2n^3 + n^2 + 4n^3}{4}$
$\displaystyle = \frac{n^4 + 2n^3 + n^2}{4}$
$\displaystyle = \frac{n^2(n^2 + 2n + 1)}{4}$
$\displaystyle = \frac{n^2(n + 1)^2}{4}$
which is your RHS.
Thus it is true for k = 2, so it true for k = 3, etc.
-Dan
Hello again, hats_06!
I must assume you know the routine for an inductive proof.
Verify $\displaystyle S(1)\!:\;\;0^3 \:=\:\left[\frac{1(0)}{2}\right]^2 \:=\:0$ . . . True!Prove by induction: .$\displaystyle 0^3 + 1^3 + 2^3 + 3^3 + \cdots + (n-1)^3 \;=\;\left[\frac{n(n-1)}{2}\right]^2$
Assume $\displaystyle S(k)$ is true: .$\displaystyle 0^3 + 1^3 + 2^3 + 3^3 + \cdots + (k-1)^3 \:=\:\left[\frac{k(k-1)}{2}\right]^2$
Add $\displaystyle k^3$ to both sides:
. . $\displaystyle \underbrace{1^3 + 2^3 + 3^3 + \cdots + (k-1)^3 + k^3}_{\text{This is the left side of }S(k+1)} \;=\;\left[\frac{k(k-1)}{2}\right]^2 +k^3$
The right side is: .$\displaystyle \frac{k^2(k-1)^2}{4} + k^3$
Factor: .$\displaystyle \frac{k^2}{4}\left[(k-1)^2 + 4k\right] \;=\;\frac{k^2}{4}\left[k^2-2k+1+4k\right] \;=\;\frac{k^2}{4}\left(k^2 + 2k + 1\right] $
. . . $\displaystyle = \;\frac{k^2}{4}(k+1)^2 \;=\;\underbrace{\left[\frac{k(k+1)}{2}\right]^2}_{\text{Right side of }S(k+1)} $
We have proved $\displaystyle S(k+1)\!:\;\;0^3 + 1^3 + 2^3 + \cdots + k^3 \;=\;\left[\frac{k(k+1)}{2}\right]^2$
. . The inductive proof is complete.
Ha! . . . Dan beat me to it (again) . . .
.