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
Find the sum
This topic has expert replies
- vikrantr93
- Junior | Next Rank: 30 Posts
- Posts: 14
- Joined: Thu Feb 03, 2011 11:51 pm
- Thanked: 1 times
-
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
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!
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!
- smackmartine
- Legendary Member
- Posts: 516
- Joined: Fri Jul 31, 2009 3:22 pm
- Thanked: 112 times
- Followed by:13 members
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!
Expert please comment!
-
- Legendary Member
- Posts: 1448
- Joined: Tue May 17, 2011 9:55 am
- Location: India
- Thanked: 375 times
- Followed by:53 members
Hi,raunekk wrote:But why did you do 2.S?
I mean, what was the thought process that you to that approach?
Thanks in advance!
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!