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

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 Cheers!

But why did you do 2.S? I mean, what was the thought process that you to that approach? Thanks in advance!

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!

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. Cheers!