Multiple choice

The sum of first $9$ terms of the series $\displaystyle \dfrac {1^3}{1}+\dfrac {1^3+2^3}{1+3}+\dfrac {1^3+2^3+3^3}{1+3+5}+ ....$ is

  1. $71$
  2. $96$
  3. $142$
  4. $192$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The n-th term is (1^3 + ... + n^3) / (1 + 3 + ... + (2n-1)). Sum of cubes = (n(n+1)/2)^2. Sum of odd numbers = n^2. Term = (n^2(n+1)^2/4) / n^2 = (n+1)^2 / 4. Sum = 1/4 * sum of (n+1)^2 for n=1 to 9. Sum = 1/4 * (2^2 + 3^2 + ... + 10^2) = 1/4 * (385 - 1) = 96.

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