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Yea, it sums them up like this:

$100*1.1^0 + $100*1.1^1 + $100*1.1^2 + $100*1.1^3..............$100*1.1^11 =$2,138.43.

Because the first exponent MUST be zero, and the 12th payment would be 11..............., for this:

$100*1.1^0 =$100 x 1 =$100, which is the first payment......and so on.

I think this states you start with 100 then for 11 more months add 10%     total 12 months correct?

The question states: Start with 100 then for 12 more months adds 10 %   

    so would it be   n,0,12 ?

....well ...in Alaska port bars there is usually a bell (like a ship's bell).....if you ring it you buy a round for the house......    so I drool when there is free drinks involved !

There is TWO answers posted at your other thread.....

$2-\frac{\sqrt3}{2}+\sqrt3\\ =2\qquad +\sqrt3-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3}{2}-\frac{\sqrt3}{2}\\ =2\qquad +\frac{2\sqrt3-1\sqrt3}{2}\\ =2\qquad +\frac{1\sqrt3}{2}\\ =2+\frac{\sqrt3}{2}\\$

Well, you can sum it up like this:

∑[100*1.1^n, n, 0, 11] =$2,138.43

Yes i know the answer, i was wondering if there was a sigma equation to show this

Well, there is a financial formula for that: FV=P{[1 + R]^N - 1/ R},     Where R=Interest rate per period, N=number of periods, P=periodic payment, FV=Future value. 

FV = $100 x {[1 + 10%]^12 - 1 / R}

FV = $100 x {[1 + 0.10]^12 -1 / 0.10}

FV = $100 x {[1.10]^12 - 1 / 0.10}

FV = $100 x {[3.138428376721 - 1 / 0.10}

FV = $100 x {2.138428376721 -1/ 0.10}

FV = $100 x       21.38428376........

FV = $2,138.43                                            

7-4√3

8)

we will consider 1,2,3,4 and 2,1,4,3 as different for the sake of simplicity

with a 1/6 chance of getting any particular side from one die, 1/36 chance from getting any particular 2, 1/216 for 3, and 1/1296 for 4 dice, you have very low odds for getting just what you want. lets start with 2 fair coins. a 1/4 chance for all heads, and 1/4 for all tails. this still means the chance of getting more heads is 1/4 because only one combo has more heads than tails

back to the dice. we will start with 2

with 2 dice, there are only 6 combos with ones and 5 of them have more ones than sixes

this puts it at a 5/36 chance to have more ones than sixes from a roll.

with all four dice its harder to calculate because you have to factor in the amount of sixes more carefully. you have to remove all the ones with at least two sixes because thats half of the dice. that rules out about 5 of the 216 options with ones. three sixes is bad too. 4 more gone. one six is okay, but if theres only a single one with it, its no good. more factors rule out more options, and i cant type long enough to find them all. hope this helps you narrow it down

I do that when I smell booze.  I drool to the tune of a different ringer.  

No speaking with the bell ring......just a lot of drooling.   If I tried to speak it would be all slurry......

OK..... How is anyone supposed to KNOW that MB???   I tried TWO possible ways.....but not the third. Please post your equations more clearly so you can get the correct answers in return   Cool?

Use parentheses or extend the radical to cover things which are supposed to be covered!  

(w(w+3))/(w-5)

i guess if you wanted it simplified thats the answer...

oh ok

Don’t worry about it. ElectricPavlov may have a high score, but he speaks dummy very well.  You should hear how he speaks when someone rings a bell.

look guest stop insulting our members

oh wow suddenly realizes i was spoken to by a 10K+ person XD i should really think about how stupidly i speak around guests cause they could be really high scoring people

ALOHA! i need answers

nah just telling people the benefits of doing it just this one time. i am not a dummy. 

lol not quite a year since last october on the 26th. oh wait its the 26th of february right now. ILLUMINATI CONFIRMED XD