Question

\(S=\frac{101}{120}+\frac{1}{2.6}+\frac{1}{4.9}+\frac{1}{6.12}+...+\frac{1}{36.57}\)

Answer 1

Có sai đề ko bạn

Answer 2

\(S=\frac{101}{102}+\frac{1}{1.2.2.3}+\frac{1}{2.3.2.3}+\frac{1}{3.4.2.3}+...+\frac{1}{17.18.2.3}=\frac{101}{102}+\frac{1}{6}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)

Đặt BT trong ngoặc đơn là A

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{18-17}{17.18}=1-\frac{1}{18}=\frac{17}{18}\)

\(S=\frac{101}{120}+\frac{1}{6}.\frac{17}{18}\)

Answer 3

Nhầm

\(\frac{1}{36.57}=\frac{1}{18.19.2.3}\) Rồi làm lại như trên