Omg, another math question undoable by the not-so-geniune-moi!

What is with this :mad: ?!

Q1
1 = 1 = 2^1-1
1 + 2 = 3 = 2^2-1
1 + 2 +2^2 = 7 = 2^3-1

Assuming that the pattern continues indefinitely, find the sum of:
1 + 2 + 2^2 + 2^3+....+ 2^19 + 2^20.

Q2
A and B are two cyclits who intend to travel the same journey. A leaves at 12 noon and averages 20km/h and B leaves 45minutes later and average 25km/h. How far does B have to travel to catch up to A.

Q1 can be written as:
2^0 = 1 = 2^1 - 1
2^0 + 2^1 = 3 = 2^2 - 1
2^0 + 2^1 + 2^2 = 7 = 2^3 - 1

Following this pattern, we have
2^0 + 2^1 + ... + 2^(n-1) + 2^n = 2^(n+1) - 1

So the sum for n = 20 is 2^21 - 1, or 2,097,151

Q2:
DistanceA = 20 * t / 60
DistanceB = 25 * (t - 45) / 60
Where t is the number of minutes past noon.

B catches up to A when DistanceB = DistanceA. Solve for t, then substitute to find the distance.

Don't know anything about Q1, but Q2 is quite easy.

Given that train A left at noon and traveled 45 minutes before train B left, it had a 15K head start -- 20kph for 45 minutes gives us 5k every 15 minutes for a total 5k in 45 minutes. .... Instead of using Noon as 0 time, use 12:45 as the 0 time. Now think of it in this terms: Train A and Train B set out at the same time, with Train A 15k down the line traveling at 20kph, and Train B traveling at 25kph. This means that Train B will gain 5k on Train A every hour, therefore it will take 3 hours to make up the 15k difference. Since Train B traveled 25kph for three hours, the answer is 75k.

-tg

for the first one, Logophobic is right about the pattern, but to find the sum of the series, try the Geometric Progression.

the GP looks like 2^0, 2^1, 2^2, 2^3 and so on.
here the first term is 2^0 or 1 and the Common Ratio is 2.

so the sum of the first n^th terms of GP is: [a*(1 - rⁿ)] / (1 - r)

where a =first term, and
.........r = common ratio.

so according to your question, sum till 20^th term will be:

Code:

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2^0*(1 - 2^21) >>21 because the first term starts from 0, so 0 to 20 = 21
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