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|dw:1322004729461:dw|

Determine the general equation of ink length for Figure n.

thinking...
looks like at each step you add 5n and remove 2(n-1)

This question is from today's Canadian Intermediate Mathematics Contest

all sides are equal of that pentagon or not?

yes it's a regular pentagon

sorry I am not familiar with terminology in english

ok, I think it is:\[\frac{5n(n+1)}{2}-n(1-n)\]

sorry - I think it should be "+" after the fraction

\[\frac{5n(n+1)}{2}+n(1-n)\]

My answer is
\[5+2(n-1)+\frac{3(n+2)(n-1)}{2}\]

mine simplifies to:\[\frac{n(3n+7)}{2}\]

it is basically the sum of two series:
1) 5, 5+10, 5+10+15, ...
2) 0, 0-2, 0-2-4, ...

Why is it 5n?

My approach is
f(n) = f(n-1) + 3n + 2

5+2(n-1) + 3/2 (n+2)(n-1)

@moneybird - your answer also simplifies to the same result :-)

yeah all resulst are equivalent :D

yeah so i got it correct on the contest!

we're ALL geniuses! :=)

what grade contest is it?

Grade 8,9, and 10

I am still in Grade 10?

I guess even in mathematics - "all roads lead to Rome"!

LOL I like that quote

thanks for posing the question @moneybird - I needed some food for my brain before going to bed :-)