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

• 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
• 11 years agoHelpfull: Yes(19) No(10)
• 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)
• 11 years agoHelpfull: Yes(13) No(9)
• 2013*2^2013
• 11 years agoHelpfull: Yes(0) No(0)
• Ans 2009 it is an arithmetic geometric series
• 11 years agoHelpfull: Yes(0) No(2)
• 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)
• 11 years agoHelpfull: Yes(0) No(0)