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

• plz del it its another answer of ques
• 9 years agoHelpfull: Yes(2) No(5)
• 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
• 9 years agoHelpfull: Yes(2) No(41)
• 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.
  
  
  
• 9 years agoHelpfull: Yes(2) No(8)
• WHAT WE NEED TO FIND WITH THIS?
  
• 9 years agoHelpfull: Yes(1) No(1)
• 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
• 9 years agoHelpfull: Yes(1) No(9)
• ans is 2012*(2^2012)-2.......
• 9 years agoHelpfull: Yes(1) No(0)
• 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.)
• 9 years agoHelpfull: Yes(1) No(2)
• take diss series as s then calculate 2s then subtract s-2s u will get a gp............ answer is 2+2^2013(2011)
• 9 years agoHelpfull: Yes(1) No(1)
• 0*1+1=1
  1*2 + 2 =4
  4*3 + 3 =15
  15*4 + 4 =64
  64*5 + 5 =325
• 9 years agoHelpfull: Yes(0) No(17)
• @Mahesh
  
  ur expecting us to solve this way in an apti exam.!!!
• 9 years agoHelpfull: Yes(0) No(0)