Tính giá trị của tổng sau đây:
S= 1.2.3+2.3.5+3.5.7+...+50.51.101
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S=1.2.3+2.3.(4+1)+3.4.(5+2)+...+n(n+1)[(n+2).(n-1)=
=1.2.3+1.2.3+2.3.4+2.3.4+3.4.5+...+(n-1)n(n+1)+n(n+1)(n+2)=
=2[1.2.3+2.3.4+3.4.5+...+(n-1)n(n+1)]+n(n+1)(n+2)
Đặt
A=1.2.3+2.3.4+3.4.5+...+(n-1)n(n+1)
4A=1.2.3.4+2.3.4.4+3.4.5.4+...+(n-1)n(n+1).4=
=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+...+(n – 1).n.(n + 1).[(n + 2) – (n – 2)]
=1.2.3.4 + 2.3.4.5 – 1.2.3.4 + 3.4.5.6 – 2.3.4.5 + … + (n – 1).n(n + 1).(n + 2) – (n – 2).(n – 1).n.(n + 1)=
= (n – 1).n(n + 1).(n + 2)
2A=\(\dfrac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{2}\)
S=2A+n(n+1)(n+2)
Bài 5:
a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)
\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)
\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)
\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)
hay A=330
Vậy: A=330
a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)
\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)
\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)
\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)
\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)
A=510
B=2505
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