So sánh A = 1 + 1/(√2) + 1/(√3) + ... + 1/(√100) và B = 2√(101) - 1
so sánh A = 1 + 2^2 + 2^3 + ... + 2^99 + 2^100 và B = 2^101 -1
Ta có \(A=1+2^2+2^3+....+2^{99}+2^{100}\)
\(2A=2+2^3+2^4+2^5+...+2^{100}+2^{101}\)
Suy ra \(2A-A=2^{101}-1=B\)
Do đó A =B
Vậy A =B
A = 1 + 2^2 + 2^3 + ... + 2^99 + 2^100
2A = 2 + 2^3 + 2^4 + ... + 2^100 + 2^101
2A - A = ( 2 + 2^3 + 2^4 + ... + 2^100 + 2^101 ) - ( 1 + 2^2 + 2^3 + ... + 2^99 + 2^100 )
A = 2^101 - 1
Vì A = 2^101 - 1 và B = 2^101 - 1
=> A = B
Vậy A=B
A=1+2^2+2^3+...+2^99+2^100
2A=2+2^3+2^4+...+2^100+2^101
2A-A=(2+2^3+2^4+...+2^100+2^101)-(1+2^2+2^3+...+2^99+2^100)
A=2^101-[2-(1+2^2)]
A=2^101-3
Vậy A=2^101-3 và B=2^101-1
=> A<B
A=1+2^1+2^2+2^3+....+2^100
B=2^101
So sánh A và B
A=1+21+22+23+...+2100
2A=2+22+23+24+...+2101
2A-A=2101-1
A=2101-1
Ta có 2101>2101-1 nên B>A
2A=2+2^2+2^3+2^4+....+2^101
=> 2A-A=(2+2^2+2^3+2^4+....+2^101)-(1+2+2^2+2^3+...+2^100)
<=> A=2^101-1 > B=2^101
2A=2+2^2+...+2^101
=>2A-A=(2+2^2+...+2^101)-(1+2+2^2+...+2^100)
=> A=2^101-1<2^101=B
vậy a<b
Chứng tỏ rằng : \(5^{27}\) <\(2^{63}\) <\(5^{28}\)
So sánh
a, A=1+2+\(2^2\) +...+\(2^4\) và B=\(2^5\) -1
b, C= 3+\(3^2\) +...+\(3^{100}\) và D= \(\dfrac{3^{101}-3}{2}\)
2:
a: A=1+2+2^2+2^3+2^4
=>2A=2+2^2+2^3+2^4+2^5
=>A=2^5-1
=>A=B
b: C=3+3^2+...+3^100
=>3C=3^2+3^3+...+3^101
=>2C=3^101-3
=>\(C=\dfrac{3^{101}-3}{2}\)
=>C=D
Ta có:
\(\left\{\begin{matrix}5^{27}=\left(5^3\right)^9=125^9\\2^{63}=\left(2^7\right)^9=128^9\end{matrix}\right\}\Rightarrow5^{27}< 2^{63}\left(1\right)\)
\(\left\{\begin{matrix}2^{63}=\left(2^9\right)^7=512^7\\5^{28}=\left(5^4\right)^7=625^7\end{matrix}\right\}\Rightarrow2^{63}< 5^{28}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow5^{27}< 2^{63}< 5^{28}\) (đpcm)
\(a.5^{27}=\left(5^3\right)^9=125^9\\ 2^{63}=\left(2^7\right)^9=128^9\)
Vì 1289 > 1259 => 263 > 527
\(5^{28}=\left(5^4\right)^7=625^7\\ 2^{63}=\left(2^9\right)^7=512^7\)
Vì 6257 > 5127 = > 528 > 263
Đã CMR: \(5^{27}< 2^{63}< 5^{28}\)
\(b.A=1+2+2^2+2^3+2^4\\ 2A=2+2^2+2^3+2^4+2^5\\ 2A-A=\left(2+2^2+2^3+2^4+2^5\right)-\left(1+2+2^2+2^3+2^4+\right)\\ A=2^5-1\\ 2^5-1=2^5-1=>A=B\\ c,C=3+3^2+....+3^{100}\\ 3C=3^2+......+3^{101}\\ 3C-C=\left(3^2+...+3^{101}\right)-\left(3+...+3^{100}\right)\\ 2C=3^{101}-3\\ C=\dfrac{3^{101}-3}{2}\\ \dfrac{3^{101}-3}{2}=\dfrac{3^{101}-3}{2}=>C=D\)
So sánh : A=1+7+7^2+7^3+...............+7^100 và B=7^101
A= 1+7+7^2+7^3+.....+7^100
=) 7A= 7+7^2+7^3+7^4+.....+7^101
=)7A-A=6A=7^101-1
Ta có: 7^101-1 <7^101 =) 6A<B =) A<B
tính
1+5^2+5^4+5^6+...+5^200
GIÚP GIÙM ĐI MÌNH ĐANG CẦN GẤP LẮM
AI NHANH DUNG MINH H CHO
Toán lớp 7
Anh Mai 25/12/2015 lúc 11:46
Đặt A= 1+5^2+5^4+5^6+...+5^200
=> 25A= 5^2+...+5^202
=>25A-A=(5^2+..+5^202)-(1+5^2+..+5^200)
24A=5^202-1
=>
a) 2n-3 chia hết cho n+1 tìm n
b) so sánh
A= 2^0 +2^1 +2^2 +2^3 +.....+2^100 và B= 2^101
\(a,2n-3⋮n+1\)
\(\Rightarrow2n+2-5⋮n+1\)
\(\Rightarrow2\left(n+1\right)-5⋮n+1\)
\(2\left(n+1\right)⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Rightarrow n\in\left\{-2;0;-6;4\right\}\)
vậy_
\(b,A=2^0+2^1+2^2+...+2^{100}\)
\(\Rightarrow2A=2^1+2^2+2^3+...+2^{101}\)
\(\Rightarrow2A-A=2^{101}-1\text{ hay }A=2^{101}-1\)
\(2^{101}-1< 2^{101}\)
\(\Rightarrow A< 2^{101}\)
vậy_
2n - 3 chia hết cho n+ 1
=> 2n + 2 - 5 chia hết cho n + 1
=> 2(n+1 ) - 5 chia hết cho n + 1
Mà 2(n+1 ) chia hết cho n + 1
=> 5 chia hết cho n + 1
=> n + 1 thuộc Ư(5)= {1; -1 ; 5 ; -5 }
TH1 : n + 1 = 1 => n = 0
TH2 : n + 1 = -1 => n = -2
th3 : n + 1 = 5 => n = 4
TH4 : n + 1 = -5 => n = -6
=> n thuộc {0;-2;4;6 }
So sánh A=\(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{2021}\)và B=20. So sánh A và B
So sánh A=(1/2^2+1/3^2+1/4^2+...+1/100^2)+1/101 với 1
so sánh A=2022^100+1/2022^99+1 và B=2022^101+1/2022^100+1
\(\dfrac{1}{2022}\cdot A=\dfrac{2022^{100}+1}{2022^{100}+100}=1-\dfrac{99}{2022^{100}+100}\)
\(\dfrac{1}{2022}B=\dfrac{2022^{101}+1}{2022^{101}+100}=1-\dfrac{9}{2022^{101}+100}\)
2022^100+100<2022^101+100
=>-99/2022^100+100<-99/2022^101+100
=>A<B
=> A/2022 = 2022^100+1/2022^100+2022 = 1- 2021/2022^100+2022
=> B/2022 = 2022^101+1/2022^101+2022 = 1- 2021/2022^101+2022
Nhận thấy 2022^101 + 2022 > 2022^100 + 2022
=> 2021/2022^101 + 2022 < 2021/2022^100 + 2022
=> B/2022 > A/2022 => B>A
Vậy A<B
So sánh:
A=1+2+22+23+......+2100 và B= 2101
2A=2(1+2+22+23+......+2100)
2A=2+22+23+24+......+2101
TA CÓ
2A-A=2+22+23+24+......+2101-(1+2+22+23+......+2100)
A=1+2201>2201
=>A>B