Tính:
\(B=3+\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+........+\dfrac{3}{1+2+.....+100}\)
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:
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)
\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)
\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)
\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)
chứng minh rằng
a , \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{512}-\dfrac{1}{1024}\) < \(\dfrac{1}{3}\)
b , \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+...+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) < \(\dfrac{3}{16}\)
Tính A= \(\dfrac{1}{2}+\dfrac{1}{2}.\left(1+2\right)+\dfrac{1}{3}.\left(1+2+3\right)+...+\dfrac{1}{100}.\left(1+2+3+...+100\right)\)
Áp dụng \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{n}\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2n}=\dfrac{n+1}{2}\)
Vậy:
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}=\dfrac{1+2+3+...+100}{2}-1\)
\(=\dfrac{100.101}{2}-1=5049\)
Tính : \(A=-\dfrac{1}{3}+\dfrac{1}{3^2}-\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3^{101}}\)
`3A=-1+1/3-1/3^2+.....+1/3^99-1/3^100`
`=>3A+A=4A=-1-1/3^101`
`=>A=(-1-1/3^101)/4`
bài 1: tính
a) 3/4+(-5/2)+(-3/5)
b) \(\sqrt{\left(7\right)^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}\)
c)\(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}\)
a)\(\dfrac{3}{4}-\dfrac{5}{2}-\dfrac{3}{5}=\dfrac{15}{20}-\dfrac{50}{20}-\dfrac{12}{20}=-\dfrac{47}{20}\)
b) \(\sqrt{7^2}+\sqrt{\dfrac{25}{16}-\dfrac{3}{2}}=7+\sqrt{\dfrac{1}{16}}=7+\dfrac{1}{4}=\dfrac{29}{4}\)
c) \(\dfrac{1}{2}.\sqrt{100}-\sqrt{\dfrac{1}{16}+\left(\dfrac{1}{3}\right)^0}=\dfrac{1}{2}.10-\sqrt{\dfrac{1}{16}+1}=5-\sqrt{\dfrac{17}{16}}\)
Tính các tổng sau:
a) A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
b) B=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{102^2}\)
c) C=\(\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+\dfrac{3}{1+2+3+4}+...+\dfrac{3}{1+2+3+...+100}\)
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+.........+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
Mặc dù t cx k biết làm nhưng mà trẩu qá Hằng
Tính tổng S= \(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\)
S = 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ... + 1/3^99 + 1/3^100
3S = 1 +1/3 +1/3^2 +1/3^3 + ... + 1/3^98 +1/3^99
3S - S = ( 1 + 1/3 + 1/3^2 +1/^3 + ... + 1/3^98 +1/3^99 ) - ( 1/3 + 1/3^2 + 1/3^3 + 1/3^4 +... + 1/3^99 + 1/3^100 )
2S = 1 - 1/3^100
S = (1 - 1/3^100). 1/2
Tính \(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+.....+\dfrac{1}{3^{100}}\)
\(D=1+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(3D=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(3D-D=3+1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-1-\dfrac{1}{3}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...-\dfrac{1}{3^{100}}\)
\(2D=3-\dfrac{1}{3^{100}}\)
\(2D=\dfrac{3^{111}-1}{3^{100}}\)
\(D=\dfrac{3^{111}-1}{2.3^{100}}\)
bài 3: tính bằng cách thuận tiện
a) \(\dfrac{13}{50}\) + 0,09 + \(\dfrac{41}{100}\) + 0,24 b) \(9\dfrac{1}{4}\) + \(6\dfrac{2}{7}\) + \(7\dfrac{3}{5}\) + \(8\dfrac{2}{3}\) + \(\dfrac{2}{5}\) + \(\dfrac{1}{3}\) + \(\dfrac{5}{7}\) + \(\dfrac{3}{4}\)
Bài 4: so sánh các cặp phân số sau:
a) \(\dfrac{2008}{2009}\) và \(\dfrac{10}{9}\) b) \(\dfrac{1}{a-1}\) và \(\dfrac{1}{a+1}\) (a>1)
Bài 5: cho phân số \(\dfrac{15}{39}\). Tìm 1 số tự nhiên, biết rằng khi thêm số đó vào mẫu số của phân số đã cho và giữ nguyên tử số thì được phân số mới bằng \(\dfrac{3}{11}\)
giải giúp mik vs, mik cần gấp!
Bài 3
a,26/100+0,009+41/100+0,24
0,26+0,09+0,41+0,24
(0,26+0,24)+(0,09+0,41)
0,5+0,5
=1
b,9+1/4+6+2/7+7+3/5+8+2/3+2/5+1/3+5/7+3/4
(9+6+7+8)+(2/7+5/7)+(1/4+3/4)+(3/5+2/5)+(2/3+1/3)
30+1+1+1+1
=34
Bài 4,5 khó quá mik ko bít lamf^^))
Bài 5: vì \(\dfrac{3}{11}\) = \(\dfrac{3\times5}{11\times5}\) = \(\dfrac{15}{55}\)
Vậy Khi giữ nguyên tử số thì số cần thêm vào mẫu số là:
55 - 39 = 16
Đáp số: 16
Bài 4: a, \(\dfrac{2008}{2009}\) < 1; \(\dfrac{10}{9}\) > 1
\(\dfrac{2008}{2009}\) < \(\dfrac{10}{9}\)
b, \(\dfrac{1}{a+1}\) và \(\dfrac{1}{a-1}\)
Ta có: a + 1 > a - 1 ⇒ \(\dfrac{1}{a+1}\) < \(\dfrac{1}{a-1}\)
Tính tổng sau: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)