The value of I = ∑ k = 1 98 ∫ k k + 1 k + 1 x ( x + 1 ) d x , then

Step-by-step solution

I = ∑ k = 1 98 ∫ k k + 1 k + 1 x x + 1 d x = ∑ k = 1 98 k + 1 ∫ k k + 1 1 x - 1 x + 1 d x = ∑ k = 1 98 k + 1 ℓ n x - ℓ n x + 1 k k + 1 = ∑ k = 1 98 k + 1 ℓ n k + 1 - ℓ n k + 2 - ℓ n k + ℓ n k + 1 = ∑ k = 1 98 k + 1 ℓ n k + 1 - k . ℓ n k - ∑ k = 1 98 k + 1 . ℓ n k + 2 - k . ℓ n k + 1 + ∑ k = 1 98 ℓ n k + 1 - ℓ n k (Difference series) ∴ I = 99 ℓ n 99 + - 99 ℓ n 100 + ℓ n 2 + ℓ n 99 = ℓ n 2 × 99 100 100 99 ......(i) For option (ii): Now, consider 100 99 = 1 + 99 99 = 99 C 0 + 99 C 1 99 + 99 C 2 99 2 + … + 99 C 97 99 9