\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)

\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)

\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)

\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)

\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)

\(A=2.2014.\left(1-\frac{1}{2014}\right)\)

\(A=2.2014.\frac{2013}{2014}\)

\(A=\frac{2.2014.2013}{2014}\)

\(A=2.2013\)

\(A=4026\)

A=4026

A+4026

tui ra rồi, nhân 2 vô A hả

a^4+a^2+1=(a^2+1)^2-a^2=(a^2-a+1)(a^2+a+1)

\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)

\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)

\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)

\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)

\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)

\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)

\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)

de sai roi ban oi. coi lai gium

Áp dụng a/(a^4+a^2+1)=1/2.(1/(a^2-a+1)-1/(a^2+a+1)) ta được

A=1/2.(1/(1^2-1+1)-1/(1^2+1+1)+1/(2^2-2+1)-1/(2^2+2+10)+...+1/(2014^2-2014+1)-1/(2014^2+2014+1))

A=1/2.(1-1/(2014^2+2014+1))

A=-2029105/4058211

(CHẮC CHẮN ĐÚNG)

A=2029105/4058211

ở mẫu   n⁴+n²+1=(n²+n+1)(n²-n+1)

\(\frac{2n}{n^4+n^2+1}=\frac{\left(n^2+n+1\right)-\left(n^2-2+1\right)}{\left(n^2-n+1\right)\left(n^2+n+1\right)}\)

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