NTA Abhyas JEE Main2020MathematicsSequences and SeriesPractice
The sum of the series 3 + 8 + 16 + 27 + 41 . . . . . . . upto 20 terms is equal to
Options
- A4230
- B4430
- C4330
- D4500
Correct answer
B. 4430
Step-by-step solution
S n = 3 + 8 + 16 + 27 + . . . . . . . . + t n S n =             3 + 8 + 16 + . . . . . . . . + t n - 1 + t n Subtracting, we get, 0 = 3 + 5 + 8 + 11 + . . . . . . . . + t n - t n - 1 - t n ⇒ t n = 1 + 2 + 5 + 8 + 11 + . . . . . . . . . n   t e r m s = 1 + n 2 4 + n - 1 3 = 1 + n 2 3 n + 1 S n = Σ t n = Σ 1 + 3 2 Σ n 2 + 1 2 Σ n = 3 2 n n + 1 2 n + 1 6 + 1 2 n n + 1 2 + n for n = 20 ,   S = 20 × 21 × 41 4 + 20 × 21 4 + 20 = 20 × 21