Question

Write out proof in paragraph form. Include contradictions if necessary

Answer 1

If $n=1$, then

$n(n+1)(2n+1)=1\cdot2\cdot3=6$

which is of course divisible by 6.

Assume the claim holds for $n=k$. Then if $n=k+1$, we have

$(k+1)(k+2)(2k+3)=\underbrace{k(k+1)(2k+1)}+k(2k+3)+(k+2)(2k+3)+2k(k+1)$

(I use the distributive property of multiplication to extract the first term, which we've assumed is divisible by 6)

The claim holds for $n=k+1$ if

$k(2k+3)+(k+2)(2k+3)+2k(k+1)$

is also divisible by 6. With some manipulation we can express this as

$(2k+2)(2k+3)+2k(k+1)$

$(k+1)(2(2k+3)+2k)$

$(k+1)(6k+6)$

$6(k+1)^2$

which is clearly divisible by 6, so the claim is true for $n=k+1$, and this completes the proof (by induction).