## Poll

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**21 members have voted**

Quote:WizardGood problem. I had seen it before, so stayed out of it. I wouldn't have filed it as an "easy math problem" (the title of the thread).

Well, here's an easy one:

For what positive value p does 1 + 2 p + 3 p

^{2}+ 4 p

^{3}+ 5 p

^{4}+ ... = 25?

Show your work - none of this "I approximated it on Excel" stuff.

Quote:ThatDonGuyWell, here's an easy one:

For what positive value p does 1 + 2 p + 3 p^{2}+ 4 p^{3}+ 5 p^{4}+ ... = 25?

Show your work - none of this "I approximated it on Excel" stuff.

Let equation A be 25=1 +2p +3p^2 etc... (the equation as stated)

Let equation B be 25p = p + 2p^2 + 3p^3 etc.

Let C equal A minus B to get 25(1-p) = 1 + p + p^2 + p^3 + p^4 etc

Let D be p*C = p + p^2 + p^3 etc

Then C minus D is 25(1-p)^2 = 1

solve to get p=4/5

Quote:rsactuaryLet equation A be 25=1 +2p +3p^2 etc... (the equation as stated)

Let equation B be 25p = p + 2p^2 + 3p^3 etc.

Let C equal A minus B to get 25(1-p) = 1 + p + p^2 + p^3 + p^4 etc

Let D be p*C = p + p^2 + p^3 etc

Then C minus D = 25(1-p)^2 = 1

solve to get p=4/5

Correct.

Another way to do this is to note that:

1 + 2 p + 3 p

^{2}+ 4 p

^{3}+ ... = (1 + p + p

^{2}+ p

^{3}+ ...)

^{2}.

Quote:ThatDonGuyCorrect.

Another way to do this is to note that:

1 + 2 p + 3 p^{2}+ 4 p^{3}+ ... = (1 + p + p^{2}+ p^{3}+ ...)^{2}.

Yeah, it's been too long since I did stuff like this on a regular basis or I would have recognized that. I kind of got there by first principles.

Quote:GialmereThis one made a splash in India a few years back.

I count

Quote:unJonI count

18

Correct!

Your geometry teacher would be proud.

You're playing a track and field board game with your family during the lockdown. If each of the runner pawns travels the indicated number of spaces every turn, at which numbered spot will all of the runners end up next to one another?