Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(H=\frac{1}{a^2}+\frac{2}{a^3}+\frac{3}{a^4}+...+\frac{n}{a^{n+1}}\)
\(H=\frac{a^{n-1}+2.a^{n-2}+...+\left(n-1\right).a+n}{a^{n+1}}\)
\(H=\frac{1}{a^{n+1}}.\left[\left(a^{n-2}+a^{n-2}+a+1\right)+\left(a^{n-2}+a^{n-3}+...+a+1\right)+...+\left(a+1\right)+1\right]\)
Đặt \(Sn=1+a+a^2+...+a^n\)=>\(a.Sn=a+a^2+a^3+...+a^n+a^{n+1}\)
=> \(a.Sn-Sn=a^{n+1}-1\)=>\(Sn.\left(a-1\right)=a^{n+1}-1\)=>\(Sn=\frac{a^{n+1}-1}{a-1}\)
Khi đó \(H=\frac{1}{a^{n+1}}.\left[\frac{a^n-1}{a-1}+\frac{a^{n-1}-1}{a-1}+...+\frac{a^2-1}{a-1}+\frac{a-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^n+a^{n-1}+...+a+1-\left(n+1\right)}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^n+a^{n-1}+...+a+1}{a-1}-\frac{n-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^{n+1}-1}{\left(a-1\right)^2}-\frac{n-1}{a-1}\right]\)
\(H=\frac{1}{a^{n+1}}.\left[\frac{a^{n+1}}{\left(a-1\right)^2}-\frac{1}{a-1}-\frac{n+1}{a-1}\right]\)
\(H=\frac{1}{\left(a-1\right)^2}-\frac{1}{a^{n+1}.\left(a-1\right)^2}-\frac{n+1}{a^{n+1}.\left(a-1\right)}< \frac{1}{\left(a-1\right)^2}\)(đpcm)
Xong rồi đó , phù.......
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
ta có:\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}
câu 2:đặt B=1/1*2+1/2*3+...+1/2007*2008
\(A=3\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\right)\)
\(\frac{A}{3}=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2008^2}\)\( (1)
mà \(B=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2007.2008}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2007}-\frac{1}{2008}\)
\(=1-\frac{1}{2008}\)<1 (2)
mà 1<3 (3)
từ (1),(2) và (3)=> đpcm
1/
\(\frac{2n+1}{n-3}+\frac{3n-5}{n-3}-\frac{4n-5}{n-3}=\frac{2n+1+\left(3n-5\right)-\left(4n-5\right)}{n-3}=\frac{2n+1+3n-5-4n+5}{n-3}=\frac{n+1}{n-3}=\frac{n-3+4}{n-3}=\frac{n-3}{n-3}+\frac{4}{n-3}=1+\frac{4}{n-3}\)
Để S là số nguyên <=> n - 3 thuộc Ư(4) = {1;-1;2;-2;4;-4}
| n-3 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 4 | 2 | 5 | 1 | 7 | -1 |
Vậy...
a) (x-3)+(x-2)+(x-1)+....+10+11=11
(x-3)+(x-2)+(x-1)+....+10 =0
gọi số hạng của tổng vế trái là n
(x-3+10).\(\frac{n}{2}\)=0
(x+7).n:2=0
(x+7) =0
\(\Rightarrow\)x+7=0 (do n\(\ne\)0)
x=0-7
x=-7
b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
\(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)
\(\frac{11}{18}<=x<=\frac{13}{9}\)
do x\(\in\)z nên x=1
vậy x=1
câu hỏi của bạn tớ cũng đang mắc
Bạn cũng có đề này à nguyễn tiến hanh ?