If [tex]n=1[/tex], then
[tex]n(n+1)(2n+1)=1\cdot2\cdot3=6[/tex]
which is of course divisible by 6.
Assume the claim holds for [tex]n=k[/tex]. Then if [tex]n=k+1[/tex], we have
[tex](k+1)(k+2)(2k+3)=\underbrace{k(k+1)(2k+1)}+k(2k+3)+(k+2)(2k+3)+2k(k+1)[/tex]
(I use the distributive property of multiplication to extract the first term, which we've assumed is divisible by 6)
The claim holds for [tex]n=k+1[/tex] if
[tex]k(2k+3)+(k+2)(2k+3)+2k(k+1)[/tex]
is also divisible by 6. With some manipulation we can express this as
[tex](2k+2)(2k+3)+2k(k+1)[/tex]
[tex](k+1)(2(2k+3)+2k)[/tex]
[tex](k+1)(6k+6)[/tex]
[tex]6(k+1)^2[/tex]
which is clearly divisible by 6, so the claim is true for [tex]n=k+1[/tex], and this completes the proof (by induction).
