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Find the sum

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Find the sum

by vikrantr93 » Sat May 21, 2011 8:19 am
Find the sum of the series :
1.2 + 2.2^2 + 3.2^3 + ... + 100.2^100


a) 100.2^101 + 2
b) 99.2^100 + 2
c) 99.2^101 + 2
d) 100.2^100 + 2
e) None of these
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by Frankenstein » Sat May 21, 2011 10:03 am
Hi,
Let S= 1.2 + 2.2^2 + 3.2^3 + ... + 100.2^100
Now, 2.S= 1.2^2 + 2.2^3 + ... + 99.2^100 + 100.2^101
Subtracting S from 2S gives
S = -(1.2+1.2^2+1.2^3+...+1.2^100) + 100.2^101
= 100.2^101 - [2.(2^100 - 1)/(2-1)]
= 100.2^101 - [2^101-2]
= 99.2^101 + 2
Hence, answer C

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by raunekk » Sat May 21, 2011 11:08 pm
But why did you do 2.S?

I mean, what was the thought process that you to that approach?

Thanks in advance!
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by smackmartine » Sat May 21, 2011 11:15 pm
I saw this problem when I was in 10 th standard, and I remember the way it has been solved by "Frankenstein". It also involves knowledge of formula of sum of geometric series = a(1-r^n)/(1-r) , where a is the 1st term and r is the ratio by which the term progresses .Its tough to remember this formula. If this question comes on the Test day, I am sure there should be some short cut method.

Expert please comment!
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by Frankenstein » Sun May 22, 2011 1:34 am
raunekk wrote:But why did you do 2.S?

I mean, what was the thought process that you to that approach?

Thanks in advance!
Hi,
We can generalize formulas for sum of n terms of an Arithmetic Progression(AP) and Geometric Progression(GP).
But, this is a classic example of AGP: (a.b)+(a+d)(b.r)+(a+2d)(b.r^2)+....+(a+nd)(r^n)
and the approach is same for all such series. Multiply the series with the common ratio 'r' and subtract one from the other so that you will get a geometric progression + constant. As, we know the formula for summation of geometric progression, we can evaluate the desired summation.

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