Tính B= 1/1.2.3 +1/2.3.4 + 1/3.4.5 +.....+1/98/99/100
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Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)
tổng quát: với mọi số n \(\ne\) 0;ta luôn có:
\(\frac{2}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{n\left(n-1\right)}-\frac{1}{n\left(n+1\right)}\)
Đặt \(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{98.99.100}\)
\(\Rightarrow2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+......+\frac{2}{98.99.100}\)
\(\Rightarrow2S=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(\Rightarrow2S=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}=\frac{1}{1.2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)
\(\Rightarrow S=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Vậy S=4949/19800
Đề sai rồi nha bạn
Phải là:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...........+\frac{1}{98.99.100}\) chứ
Ai đồng ý với mình thì ***** nha
Bài 1:
$A=1.2+2.3+3.4+...+201.202$
$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+201.202(203-200)$
$=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+201.202.203-200.201.202$
$=(1.2.3+2.3.4+3.4.5+...+201.202.203)-(1.2.3+2.3.4+....+200.201.202)$
$=201.202.203$
$\Rightarrow A=\frac{201.202.203}{3}=2747402$
Bài 2:
$S=4.5+5.6+6.7+....+100.101$
$3S=4.5(6-3)+5.6.(7-4)+6.7.(8-5)+....+100.101(102-99)$
$=4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+....+100.101.102-99.100.101$
$=(4.5.6+5.6.7+6.7.8+...+100.101.102)-(3.4.5+4.5.6+5.6.7+...+99.100.101)$
$=100.101.102-3.4.5$
$\Rightarrow S=\frac{100.101.102-3.4.5}{3}=343380$
\(B=\dfrac{1}{2}\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{17\cdot18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{17\cdot18}-\dfrac{1}{18\cdot19}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{342}\right)=\dfrac{1}{2}\cdot\dfrac{85}{171}=\dfrac{85}{342}\)
549 + X = 1326
X = 1326 - 549
X = 777
X - 636 = 5618
X = 5618 + 636
X = 6254