Question

Tính giá trị biểu thức:

C=\(\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{2499}\right)\)

Answer 1

\(C=\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{8}\right).\left(1+\dfrac{1}{15}\right)...\left(1+\dfrac{1}{2499}\right)\)

\(C=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}...\dfrac{2500}{2499}\)

\(C=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}.\dfrac{4.4}{3.5}...\dfrac{50.50}{49.51}\)

\(C=\dfrac{2.2.3.3.4.4...50.50}{1.3.2.4.3.5...49.51}\)

\(C=\dfrac{2.3.4...50}{1.2.3...49}.\dfrac{2.3.4...50}{3.4.5...51}\)

\(C=50.\dfrac{2}{51}\)

\(C=\dfrac{100}{51}\)