Hello!
Could somebody prove this for me?
1+3+6+...+n(n+1)/2=n(n+1)(n+2)/6
I mean for n=k+1
See The Math Forum.
P(k)
$\displaystyle \displaystyle\sum_{i=1}^k\frac{i(i+1)}{2}=\frac{k( k+1)(k+2)}{6}$
P(k+1)
$\displaystyle \displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\fra c{(k+1)(k+2)(k+3)}{6}$
Proof that P(k+1) will also be true if P(k) is true
$\displaystyle \displaystyle\sum_{i=1}^{k+1}\frac{i(i+1)}{2}=\sum _{i=1}^k\frac{i(i+1)}{2}+\frac{(k+1)(k+2)}{2}=\sum _{i=1}^k\frac{i(i+1)}{2}+\frac{3(k+1)(k+2)}{6}$
which, if P(k) is valid, is
$\displaystyle \displaystyle\frac{k(k+1)(k+2)}{6}+\frac{3(k+1)(k+ 2)}{6}=\frac{(k+1)(k+2)}{6}(k+3)$