how to find the sum of series whose difference is in AP?
eg. 1, 3, 6, 10, 15, 21.....

• let s is the summation of the series and Tn is the nth term.
  S = 1 + 3 + 6 + 10 + 15 + 21 ...+ Tn ---(1).
  S = 1 + 3 + 6 + 10 + 15 ...+ T(n-1) + Tn ---(2).
  subtracting eqation (2). from (1). we have...
  0 = 1 + 2 + 3 + 4 + 5 + 6 ... +(Tn- T(n-1)) -Tn
  => Tn= 1 + 2 + 3 + 4 + 5 + 6 ... +(Tn- T(n-1))
  => Tn= 1 + 2 + 3 + 4 + 5 + 6 ... + n terms
  (according to the series apply the formula... eg. A.P. or G.P. series
  here we can use,sum of n natural no, i.e. sum=n(n+1)/2)
  => Tn= n(n+1)/2
  now,
  S = ∑Tn
  => S = ∑ (n^2+n)/2
  => S = 1/2(∑n^2 + ∑n)
  => S = 1/2( (n(n+1)(2n+1))/6 + (n(n+1))/2 )
  => S = 1/2 (n/6)[ (n+1)(2n+1) + 3n +3]
  => S = n/12(2n^2+ 3n +1+ 3n +3)
  => S =n/12(2n^2+ 6n +4)
  => S=n/12[(2n +2) (n+2)]
  => S=n/6[(n +1) (n+2)]
  
• 11 years agoHelpfull: Yes(1) No(4)