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Ta có \(A=\sum\limits^n_{k=1}k^2=\sum\limits^n_{k=1}C^1_k+2\sum\limits^n_{k=1}C^2_k\)
Kết hợp với bài 2.15 ta được :
\(A=C_{n+1}^2+2C^3_{n+1}=\dfrac{n\left(n+1\right)}{2}+\dfrac{\left(n-1\right)n\left(n+1\right)}{3}=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(Un=\dfrac{4n}{12+\left(2+n^2\right)^2}\)
\(An=\sum\limits^n_{k=1}Uk\) , Tính lim An
Em cảm ơn ạ !!!!
\(u_n=\dfrac{4n}{n^4+4n^2+16}=\dfrac{4n}{n^4+8n^2+16-4n^2}=\dfrac{4n}{\left(n^2+4\right)^2-4n^2}=\dfrac{4n}{\left(n^2-2n+4\right)\left(n^2+2n+4\right)}\)
\(=\dfrac{1}{n^2-2n+4}-\dfrac{1}{n^2+2n+4}=\dfrac{1}{\left(n-1\right)^2+3}-\dfrac{1}{\left(n+1\right)^2+3}\)
Do đó:
\(A_n=\dfrac{1}{\left(1-1\right)^2+3}-\dfrac{1}{\left(1+1\right)^2+3}+\dfrac{1}{\left(2-1\right)^2+3}-\dfrac{1}{\left(2+1\right)^2+3}+...+\dfrac{1}{\left(n-1\right)^2+3}-\dfrac{1}{\left(n+1\right)^2+3}\)
\(=\dfrac{1}{0^2+3}-\dfrac{1}{2^2+3}+\dfrac{1}{1^2+3}-\dfrac{1}{3^2+3}+\dfrac{1}{2^2+3}-\dfrac{1}{4^2+3}+...+\dfrac{1}{\left(n-1\right)^2+3}-\dfrac{1}{\left(n+1\right)^2+3}\)
\(=\dfrac{1}{0^2+3}+\dfrac{1}{1^2+3}-\dfrac{1}{n^2+3}-\dfrac{1}{\left(n+1\right)^2+3}=\dfrac{7}{12}-\dfrac{1}{n^2+3}-\dfrac{1}{\left(n+1\right)^2+3}\)
\(\Rightarrow\lim\left(A_n\right)=\dfrac{7}{12}\)
Bạn tham khảo cách làm nha
https://diendantoanhoc.org/topic/106253-lim-nto-inftyprod-k1nfrac2k-12k/
