prove by induction

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Jun 19th 2008, 01:19 AM
chhatrashal

prove by induction

plz help me with this

prove by mathematical induction
P(n):1+3+6+...........n(n+1)(n+2)/2=n(n+1)(n+2)/6
Jun 19th 2008, 03:17 AM
topsquark

Quote:

Originally Posted by chhatrashal

plz help me with this

prove by mathematical induction
P(n):1+3+6+...........n(n+1)(n+2)/2=n(n+1)(n+2)/6

It's no wonder you are having problems. It isn't true.

You have either typed it incorrectly or your original problem statement has a typo.

-Dan

Edit:
As I recall
$\sum_{i = 1}^{n}i^2 = \frac{n(n + 1)(n + 2)}{6}$

Is this your problem?
Jun 19th 2008, 04:55 AM
Krizalid

But that $n+2$ should be $2n+1.$ :)
Jun 19th 2008, 08:09 AM
Soroban

Hello, chhatrashal!

Yes, there is a typo . . .

Quote:

Prove by mathematical induction:
. . $P(n)\!:\;\;1+3+6+ \hdots + {\color{blue}\frac{n(n+1)}{2}} \;=\;\frac{n(n+1)(n+2)}{6}$

This is the sum of triangular numbers . . . and it's true.

Verify $S(1)\!:\;\;1 \:=\:\frac{1\cdot2\cdot3}{6}$ . . . True.

Assume $S(k)\!:\;\;1 + 3 + 6 + \hdots + \frac{k(k+1)}{2} \;=\;\frac{k(k+1)(k+2)}{6}$

Add $\frac{(k+1)(k+2)}{2}$ to both sides;

. . $\underbrace{1 + 3 + 6 + \hdots + \frac{(k+1)(k+2)}{2}}_{\text{The left side of }S(k+1)} \;=\;\frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2}$

The right side is: . $\frac{k(k+1)(k+2)}{6} + \frac{(k+1)(k+2)}{2} \;=\;\frac{k(k+1)(k+2)}{6} + \frac{3(k+1)(k+2)}{6}$

. . $\text{Factor: }\;\frac{(k+1)(k+2)}{6}\bigg[k + 3\bigg] \;=\;\underbrace{\frac{(k+1)(k+2)(k+3)}{6}}_{\text {The right side of }S(k+1)}$

We have established $S(k+1)$ . . . The inductive proof is complete.