Q. Find the sum of following series.
1*2+2*3+3*4+4*5+...upto n ?

• 1*2+2*3+3*4+4*5+...+n(n+1)
  Tn=n(n+1)=n^2+n
  Sum= Σ(Tn)=Σ(n^2+n)[for n=1 to n]
  =Σ(n^2)+Σ(n)[for n=1 to n]
  =n(n+1)*(2n+1)/6 + n(n+1)/2
  =n(n+1)/2 *[(2n+1)/3 + 1]
  =n(n+1)/2 *(2n+4)/3
  =n(n+1)(n+2)/3
  =1/3[n*(n+1)*(n+2)]
• 11 years agoHelpfull: Yes(7) No(0)
• nth term for this= n(n+1)
  summation of n(n+1) terms is given by
  = summation{n(n+1)}
  = summation{n^2+n}
  = summation n^2 + summation n
  = {n(n+1)(2n+1)}/6 * n(n+1)/2
  
  = n(n+1)/2 [ (2n+1)/3 + 1 ] taking common terms
  = 1/3[n(n+1)(n+2)] ans.....
• 11 years agoHelpfull: Yes(2) No(0)