Question

Tìm n: \(\frac{1}{C^3_3}+\frac{1}{C^3_4}+\frac{1}{C^3_5}+...+\frac{1}{C^3_n}=\frac{89}{30}\)

Answer 1

\(\Leftrightarrow\frac{6}{1.2.3}+\frac{6}{2.3.4}+\frac{6}{3.4.5}+...+\frac{6}{\left(n-2\right)\left(n-1\right)n}=\frac{89}{30}\)

\(\Leftrightarrow\frac{3}{1.2}-\frac{3}{2.3}+\frac{3}{2.3}-\frac{3}{3.4}+...+\frac{3}{\left(n-2\right)\left(n-1\right)}-\frac{3}{\left(n-1\right)n}=\frac{89}{30}\)

\(\Leftrightarrow\frac{3}{1.2}-\frac{3}{\left(n-1\right)n}=\frac{89}{30}\)

\(\Leftrightarrow\frac{3}{n\left(n-1\right)}=-\frac{22}{15}\) (vô lý)

Vậy ko tồn tại n thỏa mãn