Latest TCS Aptitude Question SOLUTION: 1*2+2*2^2+3*2^3+4*2^4........+2012*2^2012 find the sum of the series ?

• Placement Papers
• /
• TCS

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40023775) TCS question Keep an EYE

1*2+2*2^2+3*2^3+4*2^4........+2012*2^2012 find the sum of the series ?

Asked In TCS ANKIT SINHA (11 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (22)   (3) Submit Your Solution

Read Solution (5) :

k.s.v.santosh kumar  (11 years ago)

1*2=2
1*2+2*2^2=10
1*2+2*2^2+3*2^3=34
and so on....
clearly it is in the form of (n-1)*2^(n+1) + 2
thus for n=2012
answer is 2011*2^2013+2

Is this solution Helpfull? Yes (19) | No (10)  

chenna charan tej  (11 years ago)

the above series can be written as.... sigma n*2^n....where n=1 to 2012
sigma n*2^n can be written as sigma n * sigma 2^n....
sigma n = n(n+1)/2 = 2012*2013/2
sigma 2^n is in geometric progression...using sum of n numbers in G.p. we get = 2*(2^2012-1)...
finally ans = 2012*2013(2^2012-1)

Is this solution Helpfull? Yes (13) | No (9)  

sahana  (11 years ago)

2013*2^2013

Is this solution Helpfull? Yes | No   

S K Suman  (11 years ago)

Ans 2009 it is an arithmetic geometric series

Is this solution Helpfull? Yes | No (2)  

Shubhansh Jain  (11 years ago)

this is an arithmetico geometric series whose sum is given as:
S(n)=[a/(1-r)] + [d*r*(1-r^(n-1))/(1-r)^2] - {[a+(n-1)*d]*r^n}/(1-r)
now, in the given sequence; a=1, d=1, r=2, n=2012, putting the values we get:
S(2012)=1+2011*(2^2012)

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.