Below given are two formulae for long summation.
1) 1 + 2 + 3 + 4 + ................ + n = n(n+1)/2
2) 1^2 + 2^2 + 3^2 + 4^2 + ........ + n^2 = 1/6 n(n+1)(2n+1)
The value of
(1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100) will be

• 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
  
  = ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
  
  = (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
  
  = (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
  
  = 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
  
  = 3333
• 9 years agoHelpfull: Yes(1) No(0)
• 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
  
  = ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
  
  = (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
  
  = (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
  
  = 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
  
  = 3333
• 9 years agoHelpfull: Yes(0) No(0)
• 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
  
  = ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
  
  = (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
  
  = (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
  
  = 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
  
  = 3333
• 9 years agoHelpfull: Yes(0) No(0)
• 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
  
  = ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
  
  = (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
  
  = (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
  
  = 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
  
  = 3333
• 9 years agoHelpfull: Yes(0) No(0)
• 1% of 2 + 2% of 3 + 3% of 4 + ........ + 99% of 100
  
  = ( 1* 2 + 2* 3 + 3* 4 + ........ + 99 *100) / 100
  
  = (1/100)* [1(1+1) + 2(1+2) + 3(1+3) + ......... + 99(1+99)]
  
  = (1/100) * [(1+2+3+....+99) + (1^2 + 2^2 +3^2 +.....+ 99^2]
  
  = 1/100 * [ n(n+1)/2 + 1/6 n(n+1)(2n+1) ] put n=99
  
  = 3333
• 9 years agoHelpfull: Yes(0) No(0)