So sánh A=1/2^2+1/3^2+...+1/20^2. So sánh A với 3/4
Những câu hỏi liên quan
Cho A = 1/2×2+1/3×3+1/4×4+...+1/2012×2012
a)so sánh A với 1
b)so sánh A với 3/4
\(\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3.3}< \frac{1}{2.3}\)
......
\(\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{100.100}< \frac{1}{1.2}+..+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{2.2}+..+\frac{1}{100.100}< 1-\frac{1}{100}< 1\).Suy ra điều phải chứng minh. câu b tương tự. bấm đúng cho mình nha
Đúng 1
Bình luận (0)
Cho A=1/2*2+1/3*3+1/4*4+...+1/2011*2011
a)So Sánh A với 1
B)so sánh A với 3/4
cho a =1/3 - 2/3*2 + 3/3*3 - 4/3*4 + 5/3*5 - ...... + 2023/3*2023 - 2024/3*2024 hãy so sánh a với 20/3
Sửa đề: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)
Ta có: \(a=\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)
=>\(3a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}\)
=>\(3a+a=1-\frac23+\frac{3}{3^2}-\frac{4}{3^3}+\cdots+\frac{2023}{3^{2022}}-\frac{2024}{3^{2023}}+\frac13-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+\cdots+\frac{2023}{3^{2023}}-\frac{2024}{3^{2024}}\)
=>\(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)
Đặt \(b=-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)
=>\(3b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}\)
=>\(3b+b=-1+\frac13-\frac{1}{3^2}+\cdots-\frac{1}{3^{2022}}-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}\)
=>\(4b=-1-\frac{1}{3^{2023}}=\frac{-3^{2023}-1}{3^{2023}}\)
=>\(b=\frac{-3^{2023}-1}{4\cdot3^{2023}}\)
Ta có: \(4a=1-\frac13+\frac{1}{3^2}-\frac{1}{3^3}+\cdots-\frac{1}{3^{2023}}-\frac{2024}{3^{2024}}\)
=>\(4a=1+\frac{-3^{2023}-1}{4\cdot3^{2023}}-\frac{2024}{3^{2024}}=1+\frac{-3^{2024}-3}{4\cdot3^{2024}}-\frac{8096}{4\cdot3^{2024}}\)
=>\(4a=1-\frac{3^{2024}+8099}{4\cdot3^{2024}}=1-\frac14-\frac{8099}{4\cdot3^{2024}}=\frac34-\frac{8099}{4\cdot3^{2024}}\)
=>\(4a<\frac34\)
=>\(a<\frac{3}{16}\)
mà \(\frac{3}{16}<1<\frac{20}{3}\)
nên \(a<\frac{20}{3}\)
Đúng 0
Bình luận (0)
Cho A=40+4^1+4^2+4^3+........+4^20 Hãy so sánh 3A+1 với 63^7
So sánh A với 1/21 biết A= (1-1/2).(1-1/3).(1-1/4) . ......(1-1/19).(1-1/20)
Ta có :
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{19}\right).\left(1-\frac{1}{20}\right)\)
\(=\)\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{18}{19}.\frac{19}{20}\)
\(=\)\(\frac{1.2.3.....18.19}{2.3.4.....19.20}\)
\(=\)\(\frac{1}{20}\)
Vì \(\frac{1}{20}>\frac{1}{21}\)nên \(A>\frac{1}{21}\)
Vậy \(A>\frac{1}{21}\)
Đúng 0
Bình luận (0)
Không thực hiện phép tính hãy so sánh:a)
(
−
2
)
3
.125.32.
(
−
76
)
với
(
−
2
)
3
.125.32.
(
−
76
)
b)
(
−
1
)
.
(
−
2
)
.
(
−
3
)
...
(
−
20...
Đọc tiếp
Không thực hiện phép tính hãy so sánh:
a) ( − 2 ) 3 .125.32. ( − 76 ) với ( − 2 ) 3 .125.32. ( − 76 )
b) ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) với ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 )
a) - 2 3 .125.32. ( − 76 ) > 0 12.74 . - 3 4 . ( − 395 ) < 0 ⇒ ( − 2 ) 3 .125.32. ( − 76 ) > 12.74. ( − 3 ) 4 . ( − 395 )
b) ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > 0 ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 ) < 0 ⇒ ( − 1 ) . ( − 2 ) . ( − 3 ) ... ( − 20 ) > ( − 3 ) . ( − 4 ) . ( − 5 ) ... ( − 23 )
Đúng 0
Bình luận (0)
So sánh :A= 1/22+ 1/32+ 1/42+.............+ 1/192 với 9/20
Bài 7: Chứng tỏ rằng:1/2^2 + 1/3^2 + 1/4^2 + ...1/100^2 3/4Bài 8: So sánh A 20^10 + 1 / 20^10 - 1 và B 20^10 - 1 / 20^10 - 3.
Đọc tiếp
Bài 7: Chứng tỏ rằng:
1/2^2 + 1/3^2 + 1/4^2 + ...1/100^2 < 3/4
Bài 8: So sánh A= 20^10 + 1 / 20^10 - 1 và B= 20^10 - 1 / 20^10 - 3.
8:
\(A=\dfrac{20^{10}-1+2}{20^{10}-1}=1+\dfrac{2}{20^{10}-1}\)
\(B=\dfrac{20^{10}-3+2}{20^{10}-3}=1+\dfrac{2}{20^{10}-3}\)
mà 20^10-1>20^10-3
nên A<B
Đúng 1
Bình luận (0)
A = 1/(2^3) + 1/(3^3) + 1(4^3) +...+ 1/(21^3)
So sánh A với 1/(2^2)
Xét 3 số TN liên tiếp \(\left(n-1\right);n;\left(n+1\right)\) ta có
\(\left(n-1\right).n.\left(n+1\right)=n.\left(n^2-1\right)=n^3-n< n^3\)
\(\Rightarrow A\le\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{20.21.22}=\)
\(=\dfrac{1}{2}\left(\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{22-20}{20.21.22}\right)=\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{20.21}-\dfrac{1}{21.22}\right)=\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{21.22}\right)=\dfrac{1}{2^2}-\dfrac{1}{2.21.22}< \dfrac{1}{2^2}\)
Đúng 2
Bình luận (0)