The sum of series is equal to
- A.620
- B.660
- C.1240
- D.1860
Show answer and solution
Correct option: A
Solution
The general term of the series is (1³+2³+...+n³)/(1+2+...+n).
Using standard formulas: 1³+2³+...+n³ = [n(n+1)/2]² and 1+2+...+n = n(n+1)/2.
So each term = [n(n+1)/2]² / [n(n+1)/2] = n(n+1)/2.
The sum of 15 terms = Σ(n=1 to 15) n(n+1)/2 = (1/2)Σ(n²+n) = (1/2)[15·16·31/6 + 15·16/2] = (1/2)[1240/2... let me compute: (1/2)(1240 + 120) = (1/2)(1360) = 680...
Recalculating: Σn² = 15·16·31/6 = 1240, Σn = 120. So sum = (1/2)(1240+120) = 680.
Now subtract (1/2)(1+2+...+15) = (1/2)(120) = 60.
But the first term of the series is 1, which corresponds to n=1: 1·2/2 = 1 ✓, so the 15-term sum already includes n=1 through 15.
Result = 680 − 60 = 620.
The answer is 620 (Option 1).
Using standard formulas: 1³+2³+...+n³ = [n(n+1)/2]² and 1+2+...+n = n(n+1)/2.
So each term = [n(n+1)/2]² / [n(n+1)/2] = n(n+1)/2.
The sum of 15 terms = Σ(n=1 to 15) n(n+1)/2 = (1/2)Σ(n²+n) = (1/2)[15·16·31/6 + 15·16/2] = (1/2)[1240/2... let me compute: (1/2)(1240 + 120) = (1/2)(1360) = 680...
Recalculating: Σn² = 15·16·31/6 = 1240, Σn = 120. So sum = (1/2)(1240+120) = 680.
Now subtract (1/2)(1+2+...+15) = (1/2)(120) = 60.
But the first term of the series is 1, which corresponds to n=1: 1·2/2 = 1 ✓, so the 15-term sum already includes n=1 through 15.
Result = 680 − 60 = 620.
The answer is 620 (Option 1).
[JEE Main 2019 · 10 April · Shift II]