B=1/2+1/2^2+1/2^3+1/2^4+...+1/2^99 - 1/2^100
Những câu hỏi liên quan
bài 1
A=1*2*3+2*3*4+3*4*5+...+99*100*101
B=1*3*5+3*5*7+...+95*97*99
C=2*4+4*6+..+98*100
D=1*2+3*4+5*6+...+99*100
E=1^2+2^2+3^2+...+100^2
G=1*3+2*4+3*5+4*6+...+99*101+100*102
H=1*2^2+2*3^2+3*4^2+...+99*100^2
I=1*2*3+3*4*5+5*6*7+7*8*9+...+98*99*100
K=1^2+3^2+5^2+...+99^2
A = 1*2*3 + 2*3*4 + 3*4*5 ... + 99*100*101
=> 4A = 1*2*3*4 + 2*3*4*4 + 3*4*5*4 + ... +99*100*101*4
=> 4A = 1*2*3*4 + 2*3*4*(5 - 1) + 3*4*5*( 6 - 2) + ... + 99*100*101*(102 - 98)
=> 4A = 1*2*3*4 + 2*3*4*5 - 1*2*3*4 + 3*4*5*6 - 2*3*4*5 + ... + 99*100*101*102 - 98*99*100*101
=> 4A = 99*100*101*102
=> 4A = 101989800
=> A = 25497450
Rút gọn
A= 2^100+2^99+2^98.....+2+1
B=3^100+3^99+3^98....+3+1
C=4^100+4^99+....+4+1
D=2^100- 2^99+....+2^2 - 2 + 1
E=3^100 - 3^99 + 3^98....- 3 +1
Thu gọn
M= 2 + 2^2 + 2^3 ....+ 2^100
Cho A =2+2^2+2^3+....2^100. Tìm số tự nhiên x sao cho A + 1 = 2x
Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
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A = 1 . 2 + 2 . 3 + 3 . 4 + ......... + 98 . 99 / 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ........... + ( 1 + 2 + 3 + ...... + 98 )
B = ( 1 / 51 . 52 ) + 1 / 52 . 53 + ...... + 1 / 100 . 101 ) : ( 1 / 1 . 2 + 1 / 2 . 3 + ........ + 1 / 99 . 100 + 1 / 100 . 101
c/m
a/
1/2!+2/3!+3/4!+...+99/100!<1
b/
1*2-1/2!+2*3-1/3!+3*4-1/4!+...+99*100-1/100!<2
Tính giá trị biểu thức
a, A = (1 - 1/1+2) . (1 - 1/1+2+3) . (1- 1/1+2+3+4) . ... .(1- 1/1+2+...+100)
b, B = (2/3+ 3/4 +...+99/100).(1/2+2/3+...+98/99) - (1/2+2/3+...+99/100).(2/3+3/4+...+98/99)
c, C = \(\frac{3^3+1^3}{2^3-1^3}+\frac{5^3+2^3}{3^3-2^3}+\frac{7^3+3^3}{4^3-3^3}+...+\frac{41^3+20^3}{21^3-20^3}\)
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1/ Cho A dfrac{1}{3}-dfrac{2}{3^2}+dfrac{3}{3^3}-dfrac{4}{3^4}+.....+dfrac{99}{3^{99}}-dfrac{100}{3^{100}} Chứng minh A dfrac{3}{16}2/ Cho B(dfrac{1}{2^2}-1)(dfrac{1}{3^2}-1)....(dfrac{1}{100^2}-1) So sánh B và dfrac{-1}{2}
Đọc tiếp
1/ Cho A= \(\dfrac{1}{3}\)-\(\dfrac{2}{3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+.....+\(\dfrac{99}{3^{99}}\)-\(\dfrac{100}{3^{100}}\) Chứng minh A < \(\dfrac{3}{16}\)
2/ Cho B=(\(\dfrac{1}{2^2}\)-1)(\(\dfrac{1}{3^2}\)-1)....(\(\dfrac{1}{100^2}\)-1) So sánh B và \(\dfrac{-1}{2}\)
2:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
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1+(1+2)+(1+2+3)+...+(1+2+3+4+...+99+100)/(1*100+2*99+...+99*2+100*1)*2013
Ta chia thành hai vế (1) và (2)
Số số hạng (1) là :
( 101 - 1 ) : 1 + 1 = 101 ( số )
Tổng (1) là :
( 101 + 1 ) x 101 : 2 = 5151
Tự tính tiếp
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\(1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99+100\right)\)
\(=\left(1+1+1+...+1\right)+\left(2+2+...+2\right)+\left(3+...+3\right)+...+\left(99+99\right)+100\)
\(=1.100+2.99+3.98+...+99.2+100.1\)
Do đó kết quả của phép tính cần tìm là:
\(\frac{1.100+2.99+...+99.2+100.1}{\left(1.100+2.99+...+99.2+100.1\right).2013}=\frac{1}{2013}\)
Tính A:B
a)A=98+1/2+1/3+1/4+...+1/99
B=2/3+4/3+5/4+...+100/99
b)A=2018+1/2+1/3+1/4+...+1/2019
B=3/2+4/3+3/4+...+2020/2019
c)A=99/1+98/2+97/3+...+2/98+1/99
B=1/2+1/3+1/4+...+1/100
Giải đầy đủ
a) \(A=98+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{99}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}{\frac{3}{2}+\frac{4}{3}+...+\frac{100}{99}}=1\)
b) \(A=2018+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\)(có 2018 phân số nên ta cộng 1 vào mỗi phân số)
\(A=\left(\frac{1}{2}+1\right)+\left(\frac{1}{3}+1\right)+...+\left(\frac{1}{2019}+1\right)\)
\(A=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
Và \(B=\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}{\frac{3}{2}+\frac{4}{3}+...+\frac{2020}{2019}}=1\)
c) \(A=\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}\)
\(A=99+\frac{98}{2}+...+\frac{1}{99}\)(có 98 phân số nên ta cộng 1 vào từng phân số)
\(A=\left(\frac{98}{2}+1\right)+\left(\frac{97}{3}+1\right)+...+\left(\frac{1}{99}+1\right)+1\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}+1\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)\)
Và \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\)
\(\Rightarrow\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}}=100\)
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a)\(B=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+...+\frac{100}{99}\)
\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{99}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\right)\)
\(\Rightarrow B=98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}{98+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
b)\(B=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{2019}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right)\)
\(\Rightarrow B=2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}{2018+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}}=1.\)
Vậy \(A:B=1.\)
c)\(A=\left(1+1+...+1\right)+\frac{98}{2}+\frac{97}{3}+...+\frac{2}{98}+\frac{1}{99}\)
\(A=\left(1+\frac{98}{2}\right)+\left(1+\frac{97}{3}\right)+...+\left(1+\frac{2}{98}\right)+\left(1+\frac{1}{99}\right)\)
\(A=\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}\)
\(A=100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)\)
\(\Rightarrow A:B=\frac{A}{B}=\frac{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{98}+\frac{1}{99}}=1.\)
Vậy \(A:B=1.\)
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Tính tổng:
a) A= 1^2*2 + 2^2 *3 + 3^2*4 +...+ 99^2*100
b) B= 1*2^2 + 2*3^2 + 3*4^2 +...+ 99*100^2
c) C= 1^3 + 2^3 + 3^3 +...+ 99^3
1.Tính
B=1/2+2/2^2+3/2^3+4/2^4+.....+99/2^99+100/2^100