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Đặt \(S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+....+\frac{2013}{1+2013^2+2013^4}\)
Xét:
\(\frac{k}{k+k^2+k^4}=\frac{1}{2}\cdot\frac{k^2+k+1-k^2+k-1}{k^4+k^2+1}\)
\(=\frac{1}{2}\cdot\frac{k\left(k+1\right)+1-k\left(k-1\right)-1}{\left(k^2+1\right)^2-k^2}\)
\(=\frac{1}{2}\left[\frac{1}{k\left(k-1\right)+1}-\frac{1}{k\left(k+1\right)+1}\right]\)
Áp dụng :
\(S=\frac{1}{2}\left[\frac{1}{1\cdot0+1}-\frac{1}{1\cdot2+1}+\frac{1}{2\cdot1+1}-\frac{1}{2\cdot3+1}+.....+\frac{1}{2013\cdot2012+1}-\frac{1}{2013\cdot2014+1}\right]\)
\(=\frac{2027091}{4054183}\)
Xét \(\frac{n}{1+n^2+n^4}=\frac{n}{n^4+2n^2+1-n^2}=\frac{n}{\left(n^2+1\right)^2-n^2}=\frac{n}{\left(n^2-n+1\right)\left(n^2+n+1\right)}=\frac{1}{2}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\)
\(\Rightarrow A=\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+...+\frac{1}{2013^2-2013+1}-\frac{1}{2013^2+2013+1}\right)\)
\(A=\frac{1}{2}\left(1-\frac{1}{2013^2+2013+1}\right)=...\)
1a)
Đặt \(a^2+a+1=t\Rightarrow a^2+a+2=t+1\)
\(\Rightarrow A=t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+5\right)\)
Mà \(a>1\Rightarrow\hept{\begin{cases}a^2+a-2>0\\a^2+a+5>0\end{cases}}\forall a>1\)
Vậy A là hợp số
1b)
Ta có :
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1=....=\left(2^{1006}-1\right)\left(2^{1006}+1\right)+1\)
\(=2^{2012}-1+1=2^{2012}\)
Xét Tử số của A ta có:
\(2014+\frac{2013}{2}+\frac{2012}{3}+....+\frac{2}{2013}=1+\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+....+\left(\frac{1}{2014}+1\right)\)\(TS=\frac{2015}{2}+\frac{2015}{3}+....+\frac{2015}{2014}+\frac{2015}{2015}\)
\(TS=2015.\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)\)
\(A=\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2015}\right)}{\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2015}\right)}=2015\)
Ta có:
\(\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+..+\frac{2}{2013}+\frac{1}{2014}\)
\(=\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{2}{2013}+1\right)+\left(\frac{1}{2014}+1\right)+1\)
\(=\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2013}+\frac{2015}{2014}+\frac{2015}{2015}\)
\(=2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)\)
Do đó: \(A=\frac{2015\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}+\frac{1}{2015}}=2015\)
2) xét tử ta có
2014+2013/2+2012/3+...+2/2013+1/2014
=(1+2013/2)+(1+2012/3)+...+(1+2/2013)+(1+1/2014)+1
=2015/2+2015/3+...+2015/2013+2015/2014+2015/2015
=2015(1/2+1/3+...+1/2013+1/2014+1/2015) (1)
mà mẫu bằng 1/2+1/3+1/4+...+1/2014+1/2015 (2)
từ (1),(2)=> phân thức trên =2015