find the sum of series 2,8,24,64....upto 100 terms

• By observing the series it will appear as
  
  (2^1)*1 + (2^2)*2 + (2^3)*3 + (2^4)*4......................................... + (2^100)*100 = s (say)_____________eq(1)
  
  multiplying both side by 2
  2s = (2^2)*1 + (2^3)*2 + (2^4)*3 + (2^5)*4......................................... + (2^101)*100 _____________eq(2)
  
  Now look the terms in eq(1) in eq(2) in the order 'second and first' both contain 2^2, 'third and second' both contain 2^3 so on..
  So, subtracting eq(1) from eq(2) yield
  
  2s-s = -(2^1)*1 + [ (2^2)*(1-2) + (2^3)*(2-3) + (2^4)*(3-4) + (2^5)*(4-5)............ + (2^100)*(99-100) ] + (2^101)*100
  
  => s = -[ (2^1) + (2^2) + (2^3) + (2^4) + ......................... +(2^100)] + (2^101)*100
  
  Notice the first term (2^1) is also included inside [ ] from outside and now the terms inside [ ] truns out to be a GP,
  Applying rule of GP sum
  
  => s = - [2*(2^100 - 1) / (2-1)] + (2^101)*100
  => s = - 2^101 + 2 + (2^101)*100
  => s = (2^101)*99 +2
  
  Hope now it's should be easy enough to understood
• 8 years agoHelpfull: Yes(22) No(3)
• it's a series with (2^n)*n
  
  Let s = sum_1to100_ [(2^n)*n]...............eq(1)
  
  
  hence 2s = sum_1to100_ [{2^(n+1)}*n]..................eq(2)
  hence 2s = sum_2to100 [(2^n)*(n-1)] + {2^(100+1)}*100.................eq(3)
  
  from eq(1)
  s = (2^1)*1 + sum_2to100[(2^n)*n]................. eq(4)
  
  
  On the operation eq(3)-eq(4)
  s = (2^101)*100 - sum_2to100 [2^n] - (2^1)*1 [by collecting terms of 2^n]
  
  => s = (2^101)*100 - sum_1to100 [2^n]
  => s = (2^101)*100 - {2*(2^100 - 1)/(2-1)}
  => s = (2^101)*100 - 2^101 + 2
  => s = (2^101)*99 +2
  
  enjoy
• 8 years agoHelpfull: Yes(2) No(2)
• sanjoy can u please explain this question in some other easy way
  
• 8 years agoHelpfull: Yes(1) No(0)
• can u explain once again plz
• 8 years agoHelpfull: Yes(1) No(2)
• Satya can you plz give d options
• 8 years agoHelpfull: Yes(1) No(0)
• I have given the explanation in youtube
• 8 years agoHelpfull: Yes(1) No(0)
• (2^(position of n))*(position of n)
  2^1*1=2
  2^2*2=8
  2^3*3=24
  2^4*4=64
  ...........
  ..............
  ...........
  2^100*100=100th term
  
• 8 years agoHelpfull: Yes(1) No(1)