Tính nhanh:
5) 1/2 + 1/2^2 + 1/2^3 + 1/2^4 +...+ 1/2^100
6) 1/1.2.3 + 1/2.3.4 + ... + 1/48.49.50
7) -1/3 + 1/3^2 - 1/3^3 + ... + 1/3^100 - 1/3^101
8) 1/1.2.3 + 1/2.3.4 + ... + 1/37.38.39
Những câu hỏi liên quan
tính:
A=2^100-2^99-2^98-...-2^2-2-1
B=1/1.2.3+1/2.3.4+...+1/37.38.39
C=(1-1/3).(1-1/6).(1-1/10)....(1/1/780)
1,Tính nhanh
A=1/3+1/3^2+1/3^3+...+1/3^2007+1/3^2008
B=1/3+1/3^2+1/3^3+...+1/3^n-1+1/3^n ; n∈N*
2,Tính tổng
a,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/2006.2007.2008
b,S=1/1.2.3+1/2.3.4+1/3.4.5+..+1/n.(n+1).(n+2); n∈N*
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
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B = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{n-1}}+\frac{1}{3^n}\)
3B = \(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-2}}+\frac{1}{3^{n-1}}\)
3B - B = \(1-\frac{1}{3^n}\)
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Ta có :
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(\Leftrightarrow\)\(2A=1-\frac{1}{3^{2008}}\)
\(\Leftrightarrow\)\(2A=\frac{3^{2008}-1}{3^{2008}}\)
\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{3^{2008}}:2\)
\(\Leftrightarrow\)\(A=\frac{3^{2008}-1}{2.3^{2008}}\)
Vậy \(A=\frac{3^{2008}-1}{2.3^{2008}}\)
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Xem thêm câu trả lời
a) \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
b) \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{37.38.39}\)
a) Ta có:
3A= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\left(1\right)\)
A= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\left(2\right)\)
Lấy (1) - (2) ta được:
1-\(\dfrac{1}{3^{100}}\)
b) Ta xét:
\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{2}{1.2.3},...,\dfrac{1}{37.38}-\dfrac{1}{38.39}=\dfrac{2}{37.38.39}\)
Ta có:
2B=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+..+\dfrac{2}{37.38.39}\)
=\(\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+..+\left(\dfrac{1}{37.38}-\dfrac{1}{38.39}\right)\)
=\(\dfrac{1}{1.2}-\dfrac{1}{38.39}=\dfrac{740}{38.39}=\dfrac{370}{741}\)
Vậy \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+..+\dfrac{2}{37.38.39}\)
=\(\dfrac{370}{741}\)
Nếu bn cảm thấy mk đúng tick cho mk nhé!
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Giúp mình với
Tính tổng
a) A = 1 phần 3 + 1 phần 3 mũ 2 + 1 phần 3 mũ 3 + ...... + 1 phần 3 mũ 8
b) B = 1 phần 1.2.3 + 1 phần 2.3.4 + 1 phần 3.4.5 + ......+ 1 phần 37.38.39
c) C = 1 mũ 2 + 2 mũ 2 + 3 mũ 2 + ....... + 97 mũ 2 + 98 mũ 2
Tìm x:\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}-3x=\left(1.2.3+2.3.4+...+98.99.100\right).\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\right)\)
1) Tìm x biết:
\(\left(1-\dfrac{3}{10}-x\right):\left(\dfrac{19}{10}-1-\dfrac{2}{5}\right)+\dfrac{4}{5}=1\)
2) Tính nhanh
a)\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
b)\(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
câu b bài 2:
\(\dfrac{1^2}{1\cdot2}\cdot\dfrac{2^2}{2\cdot3}\cdot\dfrac{3^2}{3\cdot4}\cdot\dfrac{4^2}{4\cdot5}\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\)
\(=\dfrac{1}{5}\)
câu a bài 2:
\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{10\cdot11\cdot12}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}-...-\dfrac{1}{12}\)
\(=1-\dfrac{1}{12}=\dfrac{11}{12}\)
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9 tháng 8 2016 lúc 14:40
Tính:
a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
b) \(B=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
a) \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(A=\frac{1}{1.2}-\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{9900}\)
\(A=\frac{9898}{19800}.\)
Vậy :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(A=\frac{9898}{19800}:2\)
\(A=\frac{4949}{19800}.\)
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a) A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
A = \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
A = \(\frac{1}{2}.\frac{4949}{9900}\)
A = \(\frac{4949}{19800}\)
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Xem thêm câu trả lời
Tính
a) \(\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+\frac{2}{81}+\frac{2}{243}+\frac{2}{729}\)
b) \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37.38.39}\)
\(b,\)Đặt \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{37\cdot38\cdot39}\)
\(B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{37.38\cdot38}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(2B=\frac{1}{1.2}-\frac{1}{38.39}\)
\(\Rightarrow B=\frac{\left(\frac{1}{1.2}-\frac{1}{38.39}\right)}{2}=\frac{185}{741}\)
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⇒B =
2
1.2
1 −
38.39
1
=
741
( ) 18
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A=1^2+2^2+3^2+...+99^2
B=3/1.2+3/3.4+...+3/99.100
C=2/1.4+2/4.7+2/7.11+...+2/96.99
D=1/1.2.3+1/2.3.4+...+98.99.100
E=1/2+1/2^2+1/2^3+...+1/2^100
Đề bài là tính
A=12+22+...+992
2A=22+32+...+1002
2A-A=(22+32+...+1002)-(12+22+...+992)
A=1002-12
A=10000-1
A=9999
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