If I = 98∑k=1∫kk+1 x(x+1)dx, then

I=∑98k=1∫kk+1 x(x+1)dx, k<x<k+1 ∴ k+1/x>1 and k+1/x+1<1 ∴ 1/x+1<k+1/x(x+1)<1/x ∴ ∫k+1k 1/x+1 < ∫k+1k k+1/x(x+1) < ∫k+1k 1/x ∴ ln(k+2/k+1) < ∫k+1k k+1/x(x+1) < ln(k+1/k) ∴ ∑98k=1ln(k+2/k+1) < ∑98k=1∫k+1k k+1/x(x+1) < ∑98k=1ln(k+1/k) ∴ ∑98k=1ln(k+2/k+1) < I < ∑98k=1ln(k+1/k) ∴ ln50 < I < ln99 ln(1−x) = −x − x2/2 − x3/3 − x4/4 − − ln(1−x) = x + x2/2 + x3/3 + x4/4 + ∴ −ln(1−x) = (4950+..) ∴ ln50 = (4950+positive terms) ∴ ln50 > 4950 ∴ 4950 < I < ln99 So, the answer is option (C) and (D).