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For the following two (02) items: Let (6 + 10 + 14 + up to m terms ) = (1 + 3 + 5 + 7 + up to n terms ) , where m What is the relation between m and n ?

Options

  1. An^2 = m(m+1)
  2. Bn^2 = m(m+2)
  3. Cn^2 = 2m(m+1)
  4. Dn^2 = 2m(m+2)

Correct answer

D. n^2 = 2m(m+2)

Step-by-step solution

The first series is an arithmetic progression with first term a = 6 and common difference d = 4 . The sum up to m terms is given by: S_m = m 2 [2(6) + (m-1)4] S_m = m 2 [12 + 4m - 4] = m 2 [4m + 8] = 2m(m + 2) The second series is an arithmetic progression with first term a = 1 and common difference d = 2 (which is the sum of the first n odd natural numbers). The sum up to n terms is given by: S_n = n 2 [2(1) + (n-1)2] = n^2 Equating the two sums, we get: n^2 = 2m(m + 2)

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