NTA Abhyas JEE Main2020MathematicsBinomial TheoremPractice
If K = 11 C 2 + 2 10 C 2 + 9 C 2 + 8 C 2 + . . . + 2 C 2 , then the value of K 100 is equal to
Correct answer
3.85
Step-by-step solution
We know that n C 2 = n n - 1 2 ⇒ 2 n C 2 = n n - 1 = n 2 - n ⇒ 2 ∑ n = 2 10 n C 2 = ∑ n = 2 10 n 2 - n = 2 2 + 3 2 + . . + 9 2 + 10 2 - 2 + 3 + . . + 9 + 10 = 1 2 + 2 2 + . . + 9 2 + 10 2 - 1 + 2 + 3 + . . + 10 = 10 × 11 × 21 6 - 10 × 11 2 = 385 - 11 C 2 ⇒ 11 C 2 + 2 10 C 2 + 9 C 2 + . . + 2 C 2 = 385 ⇒ K 100 = 3.85