Cho dãy tính :
1 + \(\frac{1+2}{2}+\frac{1+2+3}{3}+.......+\frac{1+2+3...+199}{199}=?\)
\(Tính:1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+.....+\frac{1+2+3+.....+199}{199}\)
Tính \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+.....+\frac{1+2+3...+199}{199}\)
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)\(=1+\frac{\frac{2.3}{2}}{2}+\frac{\frac{3.4}{2}}{3}+...+\frac{\frac{199.200}{2}}{199}\)\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+...+\frac{199.200}{199.2}\)\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{200}{2}\)\(=\frac{2+3+4+...+200}{2}\)\(=\frac{\frac{200.201}{2}}{2}\)\(=\frac{200.201}{2.2}\)\(=10050\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{199}+\frac{1}{200}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{192}{2}+\frac{199}{1}}\)
Tính nhanh giúp mình nha ! NHANH HẾT MỨC CÓ THỂ NHÉ
cho A /\(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)
cho B:\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
Tính A:B
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...........................
D=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)
Tính B=\(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....+\frac{198}{2}+\frac{199}{1}\)
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tôi nghĩ bài này biến đổi đc thui chứ ko tính đc.
\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....+\frac{198}{2}+\frac{199}{1}\)
\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+....+\left(\frac{199}{1}+1\right)-199\)
\(=\frac{200}{199}+\frac{200}{198}+....+\frac{200}{1}+\frac{200}{200}-200\)
\(=200.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{199}+\frac{1}{200}\right)\)
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+.......+\frac{1+2+3+...+199}{199}=?\)= ....................
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)
\(\text{Đ}\text{ặt}S=1+\frac{1}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+199}{199}\)
\(\Rightarrow S=1+\frac{\left(2+1\right).2}{2}+\frac{\left(3+1\right)3}{3}+...+\frac{\left(199+1\right)199}{199}\)
\(S=1+\frac{2+1}{1}+\frac{3+1}{1}+...+\frac{199+1}{1}\)
\(\Rightarrow S=1+\left(3+4+...+200\right)\)
Dãy (3+4+..+200 ) có số số hạng là :
(200-3):1+1=198 ( số )
Tổng của dãy (3+4+..+200 ) là :
(200+1)x198:2=19899
=> S=1+(3+4+...+200)
=> S=1+19899
=> S=19900
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...........+\frac{1+2+3......+199}{199}\)