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

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- Jun 19th 2008, 01:19 AMchhatrashalprove 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 AMtopsquark
- Jun 19th 2008, 04:55 AMKrizalid
But that $\displaystyle n+2$ should be $\displaystyle 2n+1.$ :)

- Jun 19th 2008, 08:09 AMSoroban
Hello, chhatrashal!

Yes, there is a typo . . .

Quote:

Prove by mathematical induction:

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

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

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

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

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

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

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

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