When \[1 + 3x + 5x^2 + 7x^3 + \dots + 99x^{49}\] is divided by \(x-1\) the remainder is

\(2000\),

\(2500\),

\(3000\),

\(3500\).

The remainder theorem tells us that when a polynomial \(p(x)\) is divided by \(x-1\) then the remainder is \(p(1)\), which in this case is \[1 + 3 + 5 + 7 + \dots + 99.\] This is an arithmetic progression, and so sums to \[\frac{1}{2}\times 50(1+99) = 2500\] and the answer is (b).