cho N = 1/22+1/32+1/42+...+1/20092+1/20102
CM: N < 1
Những câu hỏi liên quan
M = 1002– 992 + 982 – 972 + … + 22 – 12;
N = (202+ 182 + 162 + … + 42 + 22) – (192 + 172 + 152 + … + 32 + 12);
P = (-1)n.(-1)2n+1.(-1)n+1.
a:
Số số hạng trong dãy M là:
(1002-12):10+1=100(số)
=>Sẽ có 50 cặp (1002;992); (982;972);....;(22;12) có hiệu bằng 10
\(M=1002-992+982-972+...+22-12\)
\(=\left(1002-992\right)+\left(982-972\right)+...+\left(22-12\right)\)
\(=10+10+...+10\)
=10*50=500
b: \(N=\left(202+182+...+42+22\right)-\left(192+172+...+32+12\right)\)
\(=\left(202-192\right)+\left(182-172\right)+...+\left(22-12\right)\)
=10+10+...+10
=10*10=100
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S=1+\(\dfrac{1}{1-2}\)+\(\dfrac{1}{1-2+3}\)+...+\(\dfrac{1}{1-2+3-4+...+n}\)
và
S=12-22+32-42+...+n2
Cho A=1/12+1/22+1/22+1/32+1/42+..........+ 1/502<2
Cho A=1/22 + 1/32 + 1/42 + ... + 1/202
Chứng tỏ rằng A < 1
Ta có 1/2.2<1/1.2
1/3.3<1/2.3
1/4.4<1/3.4
.........................
1/20.20<1/19.20
=>1/2.2+1/3.3+1/4.4+...+1/20.20<1/1.2+1/2.3+1/3.4+...+1/19.20
=>A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/19-1/20
=>A<1/1-1/20
=>A<20/20-1/20
=>A<19/20<20/20=1
=>A<1
Vậy A<1
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Tính M
A
n
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1
4
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3
A
n
3
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n
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1
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!
, biết
C...
Đọc tiếp
Tính M= A n + 1 4 + 3 A n 3 ( n + 1 ) ! , biết C n + 1 2 + 2 C n + 2 2 + 2 C n + 3 2 + C n + 4 2 = 149
A: 1/2
B: 2/3
C: 3/4
D: 4/5
7 tháng 5 2021 lúc 11:25
Cho A = 1/22 + 1/32 + 1/42 + ... + 1/92.
CMR: 2/5 < A < 8/9.
Giải:
A=1/22+1/32+1/42+...+1/92
Ta có:
1/22<1/1.2
1/32<1/2.3
1/42<1/3.4
...
1/92<1/8.9
⇒A<1/1.2+1/2.3+1/3.4+...+1/8.9
A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9
A<1/1-1/9
A<8/9
Ta có:
1/22>1/2.3
1/32>1/3.4
1/42>1/4.5
...
1/92>1/9.10
⇒A>1/2.3+1/3.4+1/4.5+...+1/9.10
A>1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10
A>1/2-1/10
A>2/5
Vậy 2/5<A<8/9 (đpcm)
Chúc bạn học tốt!
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Chứng minh rằng:
A = 1/3 + 1/32 + 1/33 + ..........+ 1/399 < 1/2
B = 3/12x 22 + 5/22 x 32 + 7/32 x 42 +............+ 19/92 x 102 < 1
C = 1/3 + 2/32 + 3/33 + 4/34 +.........+ 100/3100 ≤ 0
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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Chứng tỏ:
D= 1/22 +1/32 +1/42 +....1/102 <1
Ta thấy \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
......
\(\dfrac{1}{10^2}< \dfrac{1}{9.10}\)
hay \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{10^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\)
\(D< 1-\dfrac{1}{10}=\dfrac{9}{10}< 1\) ( đpcm )
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Ta có \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)
\(\dfrac{1}{3.3}\)<\(\dfrac{1}{2.3}\)
\(\dfrac{1}{4.4}\)<\(\dfrac{1}{3.4}\)
.........................
\(\dfrac{1}{10.10}\)<\(\dfrac{1}{9.10}\)
=>\(\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{10.10}\)\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
=> D < 1 - \(\dfrac{1}{10}\)
=>D < \(\dfrac{9}{10}\)
=> D < \(\dfrac{10}{10}\)
Vậy D < 1
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hãy chứng minh: 1/22+1/32+1/42+...+1/20212+1/20222<1
a) cho A=1/22+1/12+1/62+...+1/1002
CTR: A<1/2
b) cho P=1/22+1/32+1/42+...+1/20232
CTR: P không là số tự nhiên
c) cho C=1/32+1/52+1/72+...+1/2021+1/202322
CTR: C không là số tự nhiên
GIÚP MÌNH VỚI Ạ. MÌNH CẦN GẤP. CẢM ƠN MỌI NGƯỜI!
https://olm.vn/cau-hoi/a-cho-a12211216211002-ctr-a12-b-cho-p122132142120232-ctr-p-khong-la-so-tu-nhien-c-cho-c132152172120211.8293222842881
Cô làm rồi em nhá
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