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1. D= 1/3 + 1/3.4 + 1/3.4.5 + 1/3.4.5....n < 1/2 + 1/3.4 + 1/4.5 + ...+ 1/ n.(n-1)
=> còn lại thì bạn có thể tự chứng minh
1) Tính C
\(C=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{n-1}{n!}\)
\(=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+...+\frac{n-1}{n!}\)
\(=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\)
\(=1-\frac{1}{n!}\)
3) a) Ta có : \(P=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{200}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{100}\)
\(=\frac{1}{101}+\frac{1}{102}+....+\frac{1}{199}+\frac{1}{200}\left(đpcm\right)\)
Ta có :
\(A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+...+\frac{\left(n-1\right)n-1}{n!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4}!+\frac{1}{3!}-\frac{1}{5!}+\frac{1}{4!}-...+\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)
\(=2-\frac{1}{n!}< 2\)
Vậy ...
Lời giải:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}\)
\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{(n+1)-n}{n(n+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
Vì \(1\in\mathbb{Z}; \frac{1}{n+1}\not\in\mathbb{Z}, \forall n\in\mathbb{N}\geq 1\Rightarrow A=1-\frac{1}{n+1}\not\in\mathbb{Z}\)
Ta có đpcm.
Ta có : D = \(2\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{25}+.....+\frac{1}{n\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{n\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{n}-\frac{1}{n+1}\)
\(\Rightarrow D=1-\frac{1}{n+1}=\frac{n+1}{n+1}-\frac{1}{n+1}=\frac{n}{n+1}\)
Vậy D không phải là số nguyên (đpcm)
\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n\left(n+2\right)}\right)\)
\(D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n\left(n+2\right)}\)
\(D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n\left(n+2\right)}\)
\(D=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{\left(n+2\right)-n}{n\left(n+2\right)}\)
\(D=\frac{3}{1.3}-\frac{1}{1.3}+\frac{5}{3.5}-\frac{3}{3.5}+\frac{7}{5.7}-\frac{5}{5.7}+...+\frac{\left(n+2\right)}{n\left(n+2\right)}-\frac{n}{n\left(n+2\right)}\)
\(D=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(D=\frac{1}{1}-\frac{1}{n+2}\)
\(D=\frac{n+2}{n+2}-\frac{1}{n+2}\)
\(D=\frac{n+2-1}{n+2}\)
\(D=\frac{n+1}{n+2}\Rightarrow D\notin Z\left(dpcm\right)\)
\(Q=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(Q=1-\frac{1}{n+1}=\frac{n}{n+1}\)
gọi d là UCLN của n,(n+1) ta có:
\(\hept{\begin{cases}n⋮d\\n+1⋮d\end{cases}\Rightarrow n+1-n⋮d\Rightarrow d=1}\)
=> Q là p/s tối giãn mà n khác 0 => Q ko thuộc Z