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The sum of the first n terms of the series 1^2+2 2^2+3^2+2 4^2+5^2+2 6^2+ is n(n+1)^2 2 when n is even. When n is odd the sum is

Options

  1. A3 n(n+1) 2
  2. Bn ^2( n +1) 2
  3. Cn(n+1)^2 4
  4. D[ n ( n +1) 2 ]^2

Correct answer

B. n ^2( n +1) 2

Step-by-step solution

If n is odd then (n-1) is even sum of odd terms = (n-1) n^2 2 +n^2= n^2(n+1) 2 .

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