Latest TCS Aptitude Question SOLUTION: 1*2+ 2*2^2+ 3*2^3+ 4*2^4……………………+2012*2^2012

• Placement Papers
• /
• TCS
• /
• Category

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

(#M40038541) TCS Category question Keep an EYE

1*2+ 2*2^2+ 3*2^3+ 4*2^4……………………+2012*2^2012

Asked In TCS Anshu sharma (10 years ago)

Solved Mahesh Godasu Read Solution (10)

Is this Puzzle helpful?   (5)   (5) Submit Your Solution

Read Solution (10) :

vivek gupta  (10 years ago)

plz del it its another answer of ques

Is this solution Helpfull? Yes (2) | No (5)  

sai indrneel  (10 years ago)

take 2 outside as common
we get 2(1+2*1^2+3*1^3+.........2012*1^2012)
then 2(1+2+.....2012)
we get 8100312 as answer

Is this solution Helpfull? Yes (2) | No (41)  

Mahesh Godasu  (10 years ago)

Consider the series:
sum=1+2*r+3*r^2+4*r^3+.....+n*r^(n-1)
multiply with r for the above equation
r*sum=r+2*r^2+3*r^3+4*r^4+....+n*r^n
now subtract both
sum(1-r)=1+r+r^2+r^3+r^4+....+r^(n-1)-n*r^n
here in the example r=2
so 1+r+r^2+r^3+....+r^(n-1)=((r^n)-1)/(r-1)
now sum(1-r)= [((r^n)-1)/(r-1)]-n*r^n
sum(r-1)=n*r^n-[((r^n)-1)/(r-1)]
sum={n*r^n-[((r^n)-1)/(r-1)]}/(r-1)
sum={[(n*r^n)/(r-1)]-[((r^n)-1)/(r-1)^2]}
the above sum is for the series, sum = 1+2*r+3*r^2+4*r^3+.....+n*r^(n-1)
now multiply the sum with r to get our required series
now sum become Actual sum= r*sum=r*{[(n*(r^n))/(r-1)]-[((r^n)-1)/(r-1)^2]}
Actual sum=2*{[(2012*(2^2012))/(2-1)]-[((2^2012)-1)/(2-1)^2]}
Actual sum=2*{[2012*(2^2012)]-[(2^2012)-1]}
Actual sum=2*(2^2011)-2
Actual sum=2^2012-2 is the solution.

Is this solution Helpfull? Yes (2) | No (8)  

NIVEDA SIVAPRAKASAM  (10 years ago)

WHAT WE NEED TO FIND WITH THIS?

Is this solution Helpfull? Yes (1) | No (1)  

Atul Kumar  (10 years ago)

We can write it as
2(1+2*1^2+3*1^3+.........2012*1^2012)
then 2(1+2+.....2012) ------i)
As 1,2,......2012 are in A.P. So using n=2012, a=1, l=2012
Sum= (2012/2)(1+2012)=2025078 --------ii)
putting ii) in i) we get 4050156 as ans

Is this solution Helpfull? Yes (1) | No (9)  

shivam  (10 years ago)

ans is 2012*(2^2012)-2.......

Is this solution Helpfull? Yes (1) | No   

avishek sen  (10 years ago)

Here the formula which would work for finding the sum upto n'th term is.....

(4*n)*2^n+2

ex....

upto 0th term sum=(4*0)*2^0+2=2
upto 1st term sum=(4*1)*2^1+2=10
upto 2nd term sum=(4*2)*2^2+2=34
upto 3rd term sum=(4*3)*2^3+2=98
.
.
.
so upto 2012th term sum=(4*2012)*2^2012+2 (ans.)

Is this solution Helpfull? Yes (1) | No (2)  

Raghav gupta  (10 years ago)

take diss series as s then calculate 2s then subtract s-2s u will get a gp............ answer is 2+2^2013(2011)

Is this solution Helpfull? Yes (1) | No (1)  

vivek gupta  (10 years ago)

0*1+1=1
1*2 + 2 =4
4*3 + 3 =15
15*4 + 4 =64
64*5 + 5 =325

Is this solution Helpfull? Yes | No (17)  

Souradip Nath  (10 years ago)

@Mahesh

ur expecting us to solve this way in an apti exam.!!!

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.