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\(\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2n}\right)=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2n-1}+\frac{1}{2n}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2n}\right)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2n-1}+\frac{1}{2n}-\frac{1}{1}-\frac{1}{2}-....-\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}\left(\text{đpcm}\right)\)
a) 21.(-35)-3.(-25).7=21.(-35)-(21.(-25)=21(-35+25)=21.(-10)=-210
b)77-11(30+7)=77-11.7+11.30=77-77-330=-330
c)85(35-27)-35(85-27)=85.35-85.27-35.85+35.27=(85.35-35.85)+(35.27-85.27)=35.27-85.27=27(35-85)=27.(-50)=-1350
d)(-25).68+(-34).(-250)=-25.68+34.2.125=-25.68+68.125=68(125-25)=68.100=6800
e)125.(-61).(-2^3).(-1)^2n+1=125.(-61).(-8).(-1)=(-8.125).61=-1000.61=-61000
h) đặt S=2100-299-298-...-22-2-1=2^100-(2^99+2^98+2^97+...+2+1)=2^100-A
ta có: A=2^99+2^98+2^97+...+2+1
2A=2^100+2^99+2^98+...+2^2+2
=>2A-A=(2^100+2^99+2^98+...+2^2+2)-(2^99+2^98+2^98+...+2+1)
=>A=2^100-1
=>S=2^100-(2^100-1)=2^100-2^100+1=1
A=2^100
a là bằng -210
ý b bằng -330
ý c bằng -1350
ý d bằng 6800
ý f bằng 225
tớ làm được 5 ý , tich nha bạn
1. D = 1/1.3 + 1/3.5 + 1/5.7 + ... + 1/11.13
D = 1/1 - 1/3 +1/3 - 1/5 +...+ 1/11-1/13
D = 1 - 1/13
D = 12/13
Vì 12/13 > 1/2 => D > 1/2
2. 3A = 3/3 + 3/3^2 +...+ 3/3^8
3A = 1+ 1/3 + 1/3^7
3A - A = (1+1/3 +...+1/3^7) - (1/3 + 1/3^2 +...+ 1/3^8)
2A = 1 - 1/3^8
2A = 2186/2187
A = 2186/2187 : 2 = 2186/2187 . 1/2 = 2186/4374
Mình chỉ biết vậy thôi! Chúc bạn làm bài tốt!
(2n-3)2n+1 : 7 = 49n
(2n-3)2n+1 : 7 = 72n
(2n-3)2n+1 = 72n .7
(2n-3)2n+1 = 72n+1
=) 2n-3 = 7
=) n=5
vậy ...
chúc bn học tôt
Câu thứ nhất: 2n+1 chia hết cho n-3
2n-6+7 chia hết cho n-3
2( n-3) +7 chia hết cho n-3
Vì 2(n-3) chia hết cho n-3 nên 7 chia hết cho n-3.
Vậy n-3 \(\inƯ\left(7\right)=\left\{1;7\right\}\)
\(\Rightarrow n\in\left\{4;10\right\}\)
b: =>2n+10 chia hết cho 2n-1
=>2n-1+11 chia hết cho 2n-1
=>\(2n-1\in\left\{1;-1;11;-11\right\}\)
hay \(n\in\left\{1;0;6;-5\right\}\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}n+1\in\left\{1;7;11;13;77;91;143;1001\right\}\\n\in\left\{16;31;...\right\}\end{matrix}\right.\)
=>\(n\in\left\{76\right\}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2n-1.2n+1}=\frac{49}{99}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}=\frac{49}{99}\)
\(1-\frac{1}{2n+1}=\frac{49}{99}\)
\(\frac{1}{2n-1}=1-\frac{49}{99}\)
\(\frac{1}{2n-1}=\frac{50}{99}\Rightarrow99=50\left(2n-1\right)\)
=>99=100n-1
=>-1-99=-100n
=>-100=-100n
=>n=1
1/1.3+1/3.5+1/5.7+.....+1/(2n-1).(2n+1)=49/99
1/2.(2/1.3+2/3.5+2/5.7+.....+2/(2n-1).(2n+1))=49/99
1/2(1/1-1/3+1/3-1/5+1/5-1/7+......+1/2n-1-1/2n+1=49/99
1/2.(1/1-2n+1)=49/99
1/2n+1=49/99:1/2
1/2n+1=98/99
\(\Rightarrow\)1.98/2n+1.98=98/99
\(\Rightarrow\)2n+1.98=99
2n+1=99:98
2n+1=99/98
n+1=99/98:2
n+1=99/192
n=99/192-1
n=-97/192