Câu hỏi của Nguyễn Văn Khang

Question

Ai giúp em 2 bài này với ạ

Đoàn Đức Hà · Accepted Answer

24. Ta có: $3k^2+3k+1=k^3+3k^2+3k+1-k^3=\left(k+1\right)^3-k^3$Do đó $a_k=\frac{\left(k+1\right)^3-k^3}{\left(k^2+k\right)^3}=\frac{\left(k+1\right)^3-k^3}{k^3.\left(k+1\right)^3}=\frac{1}{k^3}-\frac{1}{\left(k+1\right)^3}$Áp dụng ta được: $P=a_1+a_2+...+a_9$$=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{2^3}-\frac{1}{3^3}+...+\frac{1}{9^3}-\frac{1}{10^3}$$=1-\frac{1}{10^3}=\frac{999}{1000}$Câu trả lời: 24. Ta có: $3k^2+3k+1=k^3+3k^2+3k+1-k^3=\left(k+1\right)^3-k^3$Do đó $a_k=\frac{\left(k+1\right)^3-k^3}{\left(k^2+k\right)^3}=\frac{\left(k+1\right)^3-k^3}{k^3.\left(k+1\right)^3}=\frac{1}{k^3}-\frac{1}{\left(k+1\right)^3}$Áp dụng ta được: $P=a_1+a_2+...+a_9$$=\frac{1}{1^3}-\frac{1}{2^3}+\frac{1}{2^3}-\frac{1}{3^3}+...+\frac{1}{9^3}-\frac{1}{10^3}$$=1-\frac{1}{10^3}=\frac{999}{1000}$

Đoàn Đức Hà · Answer

23. Ta có: $B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{\left(6^2-1\right)}.....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}$$=\frac{1.1.3.3.5.5.....\left(2n+1\right)\left(2n+1\right)}{\left(1.3\right).\left(3.5\right).\left(5.7\right).....\left[\left(2n+1\right)\left(2n+3\right)\right]}$$=\frac{\left[1.3.5.....\left(2n+1\right)\right].\left[1.3.5.....\left(2n+1\right)\right]}{\left[1.3.5.....\left(2n+1\right)\right].\left[3.5.7.....\left(2n+3\right)\right]}$$=\frac{1}{2n+3}$Câu trả lời: 23. Ta có: $B=\frac{1^2}{2^2-1}.\frac{3^2}{4^2-1}.\frac{5^2}{\left(6^2-1\right)}.....\frac{\left(2n+1\right)^2}{\left(2n+2\right)^2-1}$$=\frac{1.1.3.3.5.5.....\left(2n+1\right)\left(2n+1\right)}{\left(1.3\right).\left(3.5\right).\left(5.7\right).....\left[\left(2n+1\right)\left(2n+3\right)\right]}$$=\frac{\left[1.3.5.....\left(2n+1\right)\right].\left[1.3.5.....\left(2n+1\right)\right]}{\left[1.3.5.....\left(2n+1\right)\right].\left[3.5.7.....\left(2n+3\right)\right]}$$=\frac{1}{2n+3}$

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