Question 972828
No. of terms in each bracket follow sequence:
1,3,5,7..... (AP with common difference=2)
So no of terms in 50th bracket is the 50th term of the above sequence
i.e. N = 1+(50-1)*2
       = 99
No starting term of 50th bracket can be found by counting the no. of terms already used in earlier brackets, since nos. are consecutive
Since we have the no. of terms in each bracket given by AP 1,3,5,7....
So no of terms in 49 brackets can be found by sum of 49 terms of above AP
So,
S(49) = {{{(49/2) * (2*1 + (49-1)*2)}}}
      = {{{2401}}}
So 50th bracket starts with 2401, and contains 99 consecutive terms
So sum in the 50th bracket 
= {{{(99/2) * (2*2401 + (99-1)*1)}}}
={{{242550}}}