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Lời giải:
Gọi tổng trên là A
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{101.102.103}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{103-101}{101.102.103}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{101.102}-\frac{1}{102.103}$
$=\frac{1}{1.2}-\frac{1}{102.103}=\frac{2626}{5253}$
$\Rightarrow A=\frac{1313}{5253}$
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(B=\dfrac{1}{4}-\dfrac{1}{2\left(n+1\right)\left(n+2\right)}\)
Đặt A bằng Biểu thức trong ngoặc
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{10-8}{8.9.10}\)
\(2A=\frac{1}{2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}=\frac{1}{2}-\frac{1}{9.10}=\frac{44}{90}\)
\(A=\frac{22}{90}\)
\(x=\frac{23}{45}:A=\frac{23}{45}:\frac{22}{90}=\frac{23}{11}=2\frac{1}{11}\)
Áp dụng công thức sau mà dải:
\(\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{1}{2}\left(\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\)
B=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
={1.2.3.(4-0)+2.3.4(5-1)+3.4.5.(6-2)+...+n(n+1)(n+2)[(n+3)-(n-1)]} : 4
= [1.2.3.4+2.3.4.5+3.4.5.6+...+n(n+1)(n+2)(n+3) - 1.2.3.4 - 2.3.4.5 - 3.4.5.6 - ... - n(n+1)(n+2)(n-1)] : 4
=\(\frac{\text{ n(n+1)(n+2)(n+3) }}{4}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\\ =\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\frac{612}{1225}\\ =\frac{306}{1225}\)(mà đây là toán 6 mà :V)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow A=\frac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
TK nha!!
Lời giải:
Gọi tổng trên là A
$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{101.102.103}$
$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+...+\frac{103-101}{101.102.103}$
$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{101.102}-\frac{1}{102.103}$
$=\frac{1}{1.2}-\frac{1}{102.103}=\frac{2626}{5253}$
$\Rightarrow A=\frac{1313}{5253}$