1. ## 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

2. 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
$\displaystyle \sum_{i = 1}^{n}i^2 = \frac{n(n + 1)(n + 2)}{6}$

3. But that $\displaystyle n+2$ should be $\displaystyle 2n+1.$

4. Hello, chhatrashal!

Yes, there is a typo . . .

Prove by mathematical induction:
. . $\displaystyle P(n)\!:\;\;1+3+6+ \hdots + {\color{blue}\frac{n(n+1)}{2}} \;=\;\frac{n(n+1)(n+2)}{6}$
This is the sum of triangular numbers . . . and it's true.

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.