The sum of first 9 terms of the series is
- A.71
- B.96
- C.142
- D.192
Show answer and solution
Correct option: B
Solution
The nth term of the series is (1³+2³+...+n³)/(1+3+5+...+(2n-1)).
Numerator = [n(n+1)/2]² and denominator = n² (sum of first n odd numbers).
So Tₙ = [n²(n+1)²/4] / n² = (n+1)²/4.
Sum of 9 terms = (1/4)·Σ(n+1)² from n=1 to 9 = (1/4)·(2²+3²+...+10²) = (1/4)·(385−1) = 384/4 = 96.
Numerator = [n(n+1)/2]² and denominator = n² (sum of first n odd numbers).
So Tₙ = [n²(n+1)²/4] / n² = (n+1)²/4.
Sum of 9 terms = (1/4)·Σ(n+1)² from n=1 to 9 = (1/4)·(2²+3²+...+10²) = (1/4)·(385−1) = 384/4 = 96.
[JEE Main 2015]