Question

tính tổng các dãy số   A=1/1.6+1/6.11+1/11.16+...+1/n.(n+5)                       B= 1.2+2.3+3.4+...+n. (n+1)

Answer 1

\(A=\frac{1}{1.6}+\frac{1}{6.11}+...+\frac{1}{n\left(n+5\right)}\)

\(A=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{n\left(n+5\right)}\right)\)

\(A=\frac{1}{5}\left(\frac{6-1}{1.6}+\frac{11-6}{6.11}+...+\frac{n+5-n}{n\left(n+5\right)}\right)\)

\(A=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{n}-\frac{1}{n+5}\right)\)

\(A=\frac{1}{5}\left(1-\frac{1}{n+5}\right)\)

\(A=\frac{n+4}{5n+25}\)

Answer 2

\(B=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(3B=1.2.3+2.3.3+3.4.3+...+n\left(n+1\right).3\)

\(3B=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)

\(3B=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)\)

\(3B=n\left(n+1\right)\left(n+2\right)\)

\(B=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)