S = \(\dfrac{4}{3\times7}\) + \(\dfrac{4}{7\times11}\)+ \(\dfrac{5}{11\times15}\)+...+\(\dfrac{4}{x\left(x+4\right)}\) = \(\dfrac{664}{1995}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{15}\)+...+ \(\dfrac{1}{x}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{664}{1995}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{x+4}\) = \(\dfrac{664}{1995}\)
\(\dfrac{1}{x+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{664}{1995}\)
\(\dfrac{1}{x+4}\) = \(\dfrac{1}{1995}\)
\(x\) + 4 =1995
\(x\) = 1995 - 4
\(x\) = 1991
Phân số cuối cùng của tổng S là: \(\dfrac{4}{1991\times1995}\)
Tổng S có số số hạng là: (1991 - 3):4 + 1 = 498
Đáp số: a, \(\dfrac{4}{1991\times1995}\)
b, \(498\)
