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Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\)
\(\Rightarrow\)\(\frac{a_1}{a_2}=\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\)
\(\Rightarrow\)\(\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\right)^n\) \(\left(1\right)\)
Lại có :
\(\left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}.....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}.....\frac{a_n}{a_{n+1}}=\frac{a_1.a_2.a_3.....a_n}{a_2.a_3.a_4.....a_{n+1}}=\frac{a_1}{a_{n+1}}\) \(\left(2\right)\)
Từ (1) và (2) suy ra đpcm : \(\left(\frac{a_1+a_2+a_3+...+a_n}{a_2+a_3+a_4+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)
Chúc bạn học tốt ~
Lời giải:
Đặt $\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=t$
Áp dụng TCDTSBN:
$t=\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}$
$\Rightarrow t^n=\left[\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}\right]^n(*)$
Lại có:
$\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}....\frac{a_n}{a_{n+1}}=t.t.t....t$
$\Rightarrow \frac{a_1}{a_{n+1}}=t^n(**)$
Từ $(*)$ và $(**)$ ta có:
$\left[\frac{a_1+a_2+a_3+....+a_n}{a_2+a_3+....+a_{n+1}}\right]^n=\frac{a_1}{a_{n+1}}$ (đpcm)
Có:
a1+a2=a3+a4=...=a2015+a1=1
=>a1+a2+a3+a4+...+a2014+a2015=1007+a2015
Mà 1007+a2015=0
=>a2015=-1007.
=>a1=1--1007
a1=1008.
Chúc học tốt^^
Có:
a1+a2=a3+a4=...=a2015+a1=1
=>a1+a2+a3+a4+...+a2014+a2015=1007+a2015
Mà 1007+a2015=0
=>a2015=-1007.
=>a1=1--1007
a1=1008.
Chúc học tốt^^
Viết \(a_n=\frac{\left(-1\right)^n.n^2}{n!}+\frac{\left(-1\right)^n.n}{n!}+\frac{\left(-1\right)^n}{n!}=\frac{\left(-1\right)^n.n}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{n!}\)
\(a_n=\frac{\left(-1\right)^n.\left(n-1+1\right)}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{n!}=\left(\frac{\left(-1\right)^n}{\left(n-2\right)!}+\frac{\left(-1\right)^n}{\left(n-1\right)!}\right)+\left(\frac{\left(-1\right)^n}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{n!}\right)\)
Đặt
\(b_n=\frac{\left(-1\right)^n}{\left(n-2\right)!}+\frac{\left(-1\right)^n}{\left(n-1\right)!};c_n=\left(\frac{\left(-1\right)^n}{\left(n-1\right)!}+\frac{\left(-1\right)^n}{n!}\right)\) với \(n\ge2\)
=> \(a_n=b_n+c_n;n\ge2\)
Vậy \(a_2+a_3+...+a_{2007}=\left(b_2+b_3...+b_{2007}\right)+\left(c_2+c_3...+c_{2007}\right)\)
Tính
\(B=b_2+b_3...+b_{2007}\)
\(B=\frac{\left(-1\right)^2}{0!}+\frac{\left(-1\right)^2}{1!}+\frac{\left(-1\right)^3}{1!}+\frac{\left(-1\right)^3}{2!}+\frac{\left(-1\right)^4}{2!}+...+\frac{\left(-1\right)^{2006}}{2005!}+\frac{\left(-1\right)^{2007}}{2005!}+\frac{\left(-1\right)^{2007}}{2006!}\)
\(B=1+\frac{\left(-1\right)^{2007}}{2006!}=1-\frac{1}{2006!}\)
Tính:
\(C=c_2+c_3+...+c_{2007}=\frac{\left(-1\right)^2}{1!}+\frac{\left(-1\right)^2}{2!}+\frac{\left(-1\right)^3}{2!}+...+\frac{\left(-1\right)^{2006}}{2006!}+\frac{\left(-1\right)^{2007}}{2006!}+\frac{\left(-1\right)^{2007}}{2007!}\)
\(C=1+\frac{\left(-1\right)^{2007}}{2007!}=1-\frac{1}{2007!}\)
Tính \(a_1=\left(-1\right)^1.\frac{3}{1!}=-3\)
Vậy \(a_1+a_2+a_3+...+a_{2007}=-3+1-\frac{1}{2006!}+1-\frac{1}{2007!}=-1-\frac{1}{2006!}-\frac{1}{2007!}\)
Câu 1:
a, \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^n}{c^n}=\frac{b^n}{d^n}=\frac{a^n+b^n}{c^n+d^n}=\frac{a^n-b^n}{c^n-d^n}\)
b,Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{a}{c}=\frac{b}{d}\cdot\frac{a}{c}\Rightarrow\frac{a^2}{b^2}=\frac{ab}{cd}\)
\(\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{b}{d}\cdot\frac{b}{d}\Rightarrow\frac{ac}{cd}=\frac{b^2}{d^2}\)
\(\Rightarrow\frac{ac}{bd}=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}\left(1\right)\)
Ta lại có: \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{a}{c}\cdot\frac{b}{d}=\frac{a+b}{c+d}\cdot\frac{a+b}{c+d}\Rightarrow\frac{ab}{cd}=\left(\frac{a+b}{c+d}\right)^2\left(2\right)\)
Từ (1) và (2) => \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
Câu 2:
\(\frac{a1}{a2}=\frac{a2}{a3}=....=\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+....+a2018}\)
\(\Rightarrow\frac{a1}{a2}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(1\right)\)
\(\frac{a2}{a3}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2\right)\)
..............
\(\frac{a2017}{a2018}=\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\left(2017\right)\)
Nhân các vế (1),(2)....(2017) ta được:
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\cdot\cdot\cdot\cdot\frac{a2017}{a2018}=\frac{a1}{a2018}=\left(\frac{a1+a2+...+a2017}{a2+a3+...+a2018}\right)^{2017}\)
Vậy...
Câu 3:
\(x_2^2=x_1x_3\Rightarrow\frac{x1}{x2}=\frac{x2}{x3}\)
\(x_3^2=x_2x_4\Rightarrow\frac{x2}{x3}=\frac{x3}{x4}\)
\(x_4^2=x_3x_5\Rightarrow\frac{x3}{x4}=\frac{x4}{x5}\)
\(x_5^2=x_4x_6\Rightarrow\frac{x4}{x5}=\frac{x5}{x6}\)
Đến đây thfi làm giống câu 2