Sum the series $1^3+3^3+5^3+.......... $ to $n$ terms is
- $ n^2\left ( 2n^2-1 \right )$
- $n\left ( 2n^2-1 \right )$
- $n\left ( 2n^2+1 \right )$
- $n^2\left ( 2n^2+1 \right )$
Reveal answer
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A
Correct answer
Explanation
The series is 1^3 + 3^3 + 5^3 + ... + (2n-1)^3. The sum of the first (2n-1) cubes is [ (2n-1)2n/2 ]^2 = n^2(2n-1)^2. Subtracting the even cubes (2^3 + 4^3 + ... + (2n-2)^3) = 8 * [1^3 + 2^3 + ... + (n-1)^3] = 8 * [ (n-1)n/2 ]^2 = 2n^2(n-1)^2. Result is n^2(4n^2 - 4n + 1 - 2n^2 + 4n - 2) = n^2(2n^2 - 1).
AI explanation
The series represents the sum of cubes of the first n odd natural numbers. The formula for this sum is n^2 multiplied by the term (2n^2 - 1). By substituting n into the formula, the sum to n terms evaluates directly to n^2(2n^2 - 1). The result is n^2(2n^2 - 1).