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a, \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\)
\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.100}\)
\(A< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A< 2-\frac{1}{50}\)
\(A< 2\)
b, \(B=2+2^2+2^3+...+2^{30}\)
Ta có :\(B=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{29}+2^{30}\right)\)
\(B=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{29}\left(1+2\right)\)
\(B=2.3+2^3.3+...+2^{29}.3\)
\(B=3\left(2+2^3+...+2^{29}\right)\)chia hết cho 3(1)
Lại có\(B=\left(2+2^2+2^4\right)+...+\left(2^{28}+2^{29}+2^{30}\right)\)
\(B=2\left(1+2+4\right)+...+2^{28}\left(1+2+4\right)\)
\(B=2.7+...+2^{28}.7\)
\(B=7\left(2+...+2^{29}\right)\) chia hết cho 7 (2)
Mà (3,7)=1 (3)
Từ (1)(2)(3) => B chia hết cho 21
\(2A=2\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^{100}}\)
Đến đây tôi chịu
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(2A-A=A=\frac{1}{2}-\frac{1}{2^{100}}\)
\(A+\frac{1}{2^{100}}=\frac{1}{2}-\frac{1}{2^{100}}+\frac{1}{2^{100}}=\frac{1}{2}\)
Vậy \(A+\frac{1}{2^{100}}=\frac{1}{2}\)
A=1/1^2+1/2^2+1/3^2+........+1/50^2
1/1^2=1/2x2=1-1/2
1/3^2=1/3x3=1-1/3
....................................
1/50^2=1/50x50=1-1/50
=>A < 1/1^2+1-1/2+1/2-1/3+1/3-1/4+.............+1/49-1/50
=>A < 1+(1-1/50)<1+1=2
=> A<2
A=1/1^2+1/2^2+1/3^2+........+1/50^2
1/1^2=1/2x2=1-1/2
1/3^2=1/3x3=1-1/3
....................................
1/50^2=1/50x50=1-1/50
=>A < 1/1^2+1-1/2+1/2-1/3+1/3-1/4+.............+1/49-1/50
=>A < 1+(1-1/50)<1+1=2
=> A<2
bài 1 . d là UCLN(3n+1;2n+1)=>3n+1 :d 2n+1:d
=> 3n+1 - 2n+1: d => 6n+3-6n+2:d=>1:d=>d=1
vậy:....
2.A=14+2^3.(14)+.....+2^57.(14) :7
A=30+2^4(2+2^2+2^3+2^4)+...+2^56(2+2^2+2^3+2^4)=30.(1+2^4+..+1^56) : 15
câu 1:
gọi UWCLN(2n + 1;3n+1)=d
=>2n+1 : d =>3(2n + 1) : d
3n + 1 : d 2(3n + 1) : d
=> 1 : d => d = 1
(ĐPCM)
câu 2:
A: 7 ; A = 7.B :7
A = 15 . C :15
A<1/1x2+1/2x3+1/3x4+...+1/99x100
A<1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A<1/1-1/100
A<99/100<1
tìmn ez
a)n 1la uoccua 3
b)uoc cua n 7
Ta có: a2-(a-1)(a+1)
= a2-(a2-a+a-1)
= a2-a2+a-a+1
=1
Vậy a2-(a-1)(a+1)=1 (đpcm)