1*2+2*3+3*4+-----------+99*100

1) 330033
2)300033
3)330000
4)333000
5)330003
6)333300

• 6) 333300
  tn = n*(n+1)
  given exp is = Σtn = Σ n*(n+1) [n=1 to n=99]
  
  = Σ(n^2+n) = n*(n+1)*(2n+1)/ 6 + n*(n+1)/2
  on solving, = n*(n+1)*(n+2) / 3
  
  put n=99, we get reqd sum = 99*100*101/3 = 333300
• 9 years agoHelpfull: Yes(0) No(0)
• ans is 6) 333300
  
  explanation : 1(1+1)+2(2+1)+3(3+1)-------99(99+1)
  
  => ((1*1)+(2*2)+(3*3)+-----(99*99))+(1+2+3+-------99)
  
  => (99*100*199)/6 + 4950
  
  => 333300
  
  OR
  
  We can solve frm formula
  
  tn = n*(n+1)
  given exp is = Σtn = Σ n*(n+1) [n=1 to n=99]
  
  
  
• 9 years agoHelpfull: Yes(0) No(0)