Multiple choice

Find the sum of first $n$ terms of the series. $1^{3}+3\times2^{2}+3^{3}+3\times4^{2}+5^{3}+3\times6^{2}+....$ If $n$ is odd.

  1. $\displaystyle \frac{1}{8}(n+1)[n^{3}+7n^{2}-3n-1]$
  2. $\displaystyle \frac{n}{8}(n^{2}+4n^{2}+10n+6)$
  3. $\displaystyle \frac{1}{8}(n+1)[n^{2}+7n^{2}-3n-1]$
  4. $\displaystyle \frac{1}{8}(n+1)[n^{2}+7n^{2}-3n]$
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A Correct answer
Explanation

For odd n, the terms consist of cubes of odd numbers and 3 times the squares of even numbers. Summing these two parts and simplifying gives (n + 1)(n^3 + 7n^2 - 3n - 1)/8.

AI explanation

For odd n, the series has (n+1)/2 cubic terms and (n-1)/2 quadratic terms, which can be summed separately. The sum of the cubic terms is 1^3 + 3^3 + ... + n^3, evaluated using the standard sum of cubes formula to get (n^2)(n+1)^2 / 8. The sum of the quadratic terms is 3(2^2 + 4^2 + ... + (n-1)^2), which equals 12 times the sum of the first (n-1)/2 squares, resulting in n(n-1)(n+1) / 2. Combining these gives a total sum of (1/8)(n+1)[n^3 + 7n^2 - 3n - 1].

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