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Lời giải:
$3S_n=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+....+\frac{(n+3)-n}{n(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}$
$=\frac{1}{1.2.3}-\frac{1}{(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{1.2.3.3}-\frac{1}{3(n+1)(n+2)(n+3)}$
$\Rightarrow S_n=\frac{1}{18}-\frac{1}{3(n+1)(n+2)(n+3)}$
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\cdot\cdot\cdot\left(1-\frac{1}{n^2}\right)\)
\(\Rightarrow A=\frac{3}{4}\cdot\frac{8}{9}\cdot\cdot\cdot\frac{n^2-1}{n^2}\)
\(\Rightarrow A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\frac{\left(n-1\right)\left(n+1\right)}{n\cdot n}\)
\(\Rightarrow A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\cdot\cdot\left[\left(n-1\right)\left(n+1\right)\right]}{\left(2\cdot2\right)\cdot\left(3\cdot3\right)\cdot\cdot\cdot\left(n\cdot n\right)}\)
\(\Rightarrow A=\frac{\left[1\cdot2\cdot\cdot\cdot\cdot\cdot\left(n-1\right)\right]\cdot\left[3\cdot4\cdot\cdot\cdot\cdot\cdot\left(n+1\right)\right]}{\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot n\right)}\)
\(\Rightarrow A=\frac{1\cdot\left(n+1\right)}{n\cdot2}\)
\(\Rightarrow A=\frac{n+1}{2n}\)
A=(1-1/2^2)(1-1/3^2).....(1-1/n^2)
A=1(1/2^2-1/3^2-...-1/n^2)
......
xin lỗi bạn nha mình phải tắt máy rồi bạn cố gắng suy nghĩ tiếp nha
Biến đổi ở phân số dạng tổng quát :
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{3}{3n(n+1)(n+2)(n+3)}=\frac{3+n-n}{3n(n+1)(n+2)(n+3)}\)
\(=\frac{1}{3}\left[\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\right]\)
\(=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Áp dụng kết quả này vào bài được :
\(\frac{1}{1\cdot2\cdot3\cdot4}=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{2\cdot3\cdot4}\right],\frac{1}{2\cdot3\cdot4\cdot5}=\frac{1}{3}\left[\frac{1}{2\cdot3\cdot4}-\frac{1}{3\cdot4\cdot5}\right],...\)
\(\frac{1}{n(n+1)(n+2)(n+3)}=\frac{1}{3}\left[\frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
Cộng từng vế,ta được : \(S=\frac{1}{3}\left[\frac{1}{1\cdot2\cdot3}-\frac{1}{(n+1)(n+2)(n+3)}\right]\)
P/S : Xong
Ta có: S= \(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(3S=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+...+\frac{3}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(=\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
\(\Rightarrow S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
Vậy \(S=\frac{\frac{1}{1.2.3}-\frac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}}{3}\)
Lời giải:
\(S=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+....+\frac{1}{n}.\frac{n(n+1)}{2}\\ =1+\frac{3}{2}+\frac{4}{2}+...+\frac{n+1}{2}\\ =\frac{2+3+4+....+(n+1)}{2}=\frac{1+2+3+....+(n+1)}{2}-\frac{1}{2}\\ =\frac{(n+1)(n+2)}{4}-\frac{1}{2}\)