1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + ..... + 127.128.129 find the sum

• the 1st term in the series is (1)(1+1)(1+2)
  the 2nd term is (2)(2+1)(2+2)
  
  therefore, the nth term is (n)(n+1)(n+2)
  which when expanded gives (n^3 + 4n^2 + 4n)
  here n is 127
  Σ (n^3 + 3n^2 + 2n) [for n = 1 to n = 127]
  Σ (n^3) + 3 Σ (n^2) + 2 Σ (n) [for n = 1 to n = 127]
  after solving it we get 68153280
  ans is 68153280
• 10 years agoHelpfull: Yes(39) No(0)
• this series can be written as sum(n.(n+1).(n+2))=sum(n^3)+sum(3n^2)+sum(2n)={n(n+1)/2}^2 +3{n(n+1)(2n+1)/6} +2{n(n+1)/2}= 68153280
• 10 years agoHelpfull: Yes(19) No(0)
• @ rudra...
  summation(n(n+1)(n+2))
  =sum(n^3)+3sum(n^2)+2sum(n)
  summation of n=n(n+1)/2 1st series
  summation of n^2=n(n+1)(2n+1)/6 2nd series
  summation of n^3=(n^2(n+1)^2)/4 3rd series n=127
  hence ans=68153280
• 10 years agoHelpfull: Yes(5) No(0)
• 68153280...
  we have 3 series 1,2,3
  2,3,4'
  3,4,5
  find nth term of each individual series and multiply them to get nth term of series. use summation and put the value of n=127
  
• 10 years agoHelpfull: Yes(1) No(1)
• @Priyank Kumar
  for 1st series- sum of the series =127/2(1+127)
  for 2nd series- sum of the series =127/2(2+128)
  for 3rd series- sum of the series =127/2(3+129)
  am i right???
  if not plzzz correct me....
• 10 years agoHelpfull: Yes(0) No(4)