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áp dụng t.c dãy tỉ số bằng nhau ta có:
\(\frac{a1}{a2}=\frac{a2}{a3}=\frac{a3}{a4}=.....=\frac{an}{an+1}=\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\)
\(\frac{a1}{a2}\cdot\frac{a2}{a3}\cdot\frac{a3}{a4}\cdot...\cdot\frac{an}{an+1}=\frac{a1}{an+1}=\left(\frac{a1}{a2}\right)^n=\left(\frac{a1+a2+a3+....+an}{a2+a3+a4+...+an+1}\right)^n\)(vì từ 1 đến n có n chữ số)
=> đpcm
Theo tính chất của dãy tỉ số bằng nha, ta có :
\(\dfrac{a_1}{a_2}=\dfrac{a_2}{a_3}=.....=\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\Rightarrow\dfrac{a_1}{a_2}=\dfrac{a_1+a_2+....+a_n}{a_2+a_3+....+a_{n+1}}\)
\(\dfrac{a_2}{a_3}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
.................................
\(\dfrac{a_n}{a_{n+1}}=\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\)
\(\Rightarrow\left(\dfrac{a_1+a_2+.....+a_n}{a_2+a_3+.....+a_{n+1}}\right)^n=\dfrac{a_1}{a_2}.\dfrac{a_2}{a_3}........\dfrac{a_n}{a_{n+1}}\)
Vậy \(\left(\dfrac{a_1+a_2+......+a_n}{a_2+a_3+......+a_{n+1}}\right)=\dfrac{a_1}{a_{n+1}}\) (đpcm)
~ Học tốt ~
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=....=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=....=\left(\frac{a_n}{a_{n+1}}\right)^n=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)(1)
Ta có: \(\left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}.\frac{a_1}{a_2}.\frac{a_1}{a_2}....\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}.\frac{a_3}{a_4}....\frac{a_n}{a_{n+1}}=\frac{a_1}{a_{n+1}}\)(2)
Từ (1), (2) \(\Rightarrow\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n=\frac{a_1}{a_{n+1}}\)(đpcm)
\(\text{Áp dụng tính chất của dãy tỉ số bằng nhau có:}\)
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\left(\frac{a_1}{a_2}\right)^n=\left(\frac{a_2}{a_3}\right)^n=...=\left(\frac{a_n}{a_{n+1}}\right)^n\)\(=\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)
Mà\( \left(\frac{a_1}{a_2}\right)^n=\frac{a_1}{a_2}\cdot\frac{a_1}{a_2}\cdot...\cdot\frac{a_1}{a_2}\)\(=\frac{a_1}{a_2}\cdot\frac{a_2}{a_3}\cdot...\cdot\frac{a_n}{a_{n+1}}\)\(=\frac{a_1}{a_{n-1}}\)
\(\Rightarrow\)\(\left(\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\right)^n\)\(=\frac{a_1}{a_{n-1}}\)
Ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_n}{a_{n+1}}=\frac{a_1+a_2+...+a_n}{a_2+a_3+...+a_{n+1}}\)
\(\Rightarrow\)\(\frac{a_1^n}{a_2^n}=\frac{a_2^n}{a_3^n}=...=\frac{a_n^n}{a_{n+1}^n}=\frac{a_1^n+a_2^n+...+a_n^n}{a_2^n+a_3^n+...+a_{n+1}^n}=\frac{\left(a_1+a_2+...+a_n\right)^n}{\left(a_2+a_3+...+a_{n+1}\right)^n}=\frac{a_1.a_2...a_n}{a_2.a_3...a_{n+1}}=\frac{a_1}{a_{n+1}}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=k\)
=>\(\frac{a_1}{a_2}.\frac{a_2}{a_3}.....\frac{a_{n-1}}{a_n}.\frac{a_n}{a_1}=k.k.....k.k\)
=>\(k^n=\frac{a_1.a_2.....a_{n-1}.a_n}{a_2.a_3.....a_n.a_1}\)
=>\(k^n=1=1^n\)
=>k=1
=>\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=1\)
=>\(a_1=a_2=...=a_n\)
\(=>\frac{a^2_1+a^2_2+...+a_n^2}{\left(a_1+a_2+...+a_n\right)^2}\)
=\(\frac{a^2_1+a^2_1+...+a_1^2}{\left(a_1+a_1+...+a_1\right)^2}\)
=\(\frac{n.a^2_1}{\left(n.a_1\right)^2}=\frac{n.a_1^2}{n^2.a^2_1}=\frac{1}{n}\)
thế này dc ko
Áp dụng t/c của dãy tỉ số bằng nhau, ta có :
\(\frac{a_1}{a_2}=\frac{a_2}{a_3}=...=\frac{a_{n-1}}{a_n}=\frac{a_n}{a_1}=\frac{a_1+a_2+...+a_{n-1}+a_n}{a_2+a_3+...+a_n+a_1}\Rightarrow a_1=a_2=...=a_n\)
\(\frac{a^1_2+a^2_2+...+a^2_n}{\left(a_1+a_2+...+a_n\right)}=\frac{na^2_1}{\left(na_1\right)^2}=\frac{1}{n}\)
dài dòng thế, nhìn lóa cả mắt
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 \(a_1+a_2+a_3+...+a_{2007}=-3+1-\frac{1}{2006!}+1-\frac{1}{2007!}=-1-\frac{1}{2006!}-\frac{1}{2007!}\)