Question

[Nguyễn Quang Tâm]

Giúp tui với, chiều tui phải nộp rồi:

Bài 10:

Tính B=\(\dfrac{5}{2.1}+\dfrac{4}{1.11}+\dfrac{3}{11.2}+\dfrac{1}{2.15}+\dfrac{13}{15.4}\)

Bài 12:Rút gọn viểu thức:

B=\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)

(Kí hiệu:n!(đọc là n giai thừa)n! là tích của n số tự nhiên liên tiếp từ 1 đến n; n!=1.2.3....n)

Bài 13:Rút gọn biểu thức sau:

C=\(\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+\dfrac{7}{\left(3.4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

Answer 1

[Khiêm Nguyễn Gia]

Bài \(13\):
\(C=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
\(=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{n^2+2n+1-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{4-1}{1\cdot4}+\dfrac{9-4}{4\cdot9}+\dfrac{16-9}{9\cdot16}+...+\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}\)
\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)
\(=1-\dfrac{1}{\left(n+1\right)^2}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Answer 2

[Khiêm Nguyễn Gia]

Bài \(10\):
\(B=\dfrac{5}{2\cdot1}+\dfrac{4}{1\cdot11}+\dfrac{3}{11\cdot2}+\dfrac{1}{2\cdot15}+\dfrac{13}{15\cdot4}\)
\(=7\left(\dfrac{5}{2\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{1}{14\cdot15}+\dfrac{13}{15\cdot28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{28}\right)\)
\(=7\left(\dfrac{1}{2}-\dfrac{1}{28}\right)=7\cdot\dfrac{13}{28}=\dfrac{13}{4}\)