I know I'm just overthinking this and it's literally probably an x=y formula but humor me. Midas wants to buy all the land near his house. There are 65 lots surrounding his home, and each lot costs $7200. But for every lot purchased, the price doubles. How much money does Midas need to buy all 65 lots?

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I know I'm just overthinking this and it's literally probably an x=y formula but humor me. Midas wants to buy all the land near his house. There are 65 lots surrounding his home, and each lot costs $7200. But for every lot purchased, the price doubles. How much money does Midas need to buy all 65 lots?

Mathematics
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Use sum of GP series here
a(R^n-1) / R-1 here a= 7200 n=65 and r=2
it says for every lot purchased, the price doubles you can try without the fancy geometric series by using simple arithmetic first Price per lot = 7200 number of lots 65 normally it would be 65* 7200 but the condition is that the price increases for every additional lot purchased so if we start with the first lot \(L_1 = 7200\) then second lot \(L_2 = 2 \times L_1\) then third \(L_3 = 2 \times L_2 \) we can quickly see that we are doubling the previous price

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so maybe it would be safe to say that our addition of lot sequence would be \(\large L_1 + L_2 + L_3 ... L_{65}\) we will worry about inserting each of \(L_n \) prices later and focus on how we can write L into condensed form. Have you done any summations in the past?
humour me
Let's pretend for a second there are only 3 lots. Then the prices for buying will be: \[7200 + 2*7200+ 2*2*7200\] Which simplifies to \[7200(1+2+4)\] So this is where the geometric series comes in, instead of only having 3, we have 65 which will be a geometric series in 2 which is exactly what @arindameducationusc is referring to here.
Nice @Kainui & @nincompoop... Good job!

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