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Grace [21]
3 years ago
7

Write out proof in paragraph form. Include contradictions if necessary

Mathematics
1 answer:
Sophie [7]3 years ago
4 0

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).

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