1+(1+2)/2+(1+2+3)/3+...(1+...199)/199=
Những câu hỏi liên quan
1+1+2/2+1+2+3/3+...+1+2+3+...+199/199=
1/2+1+1/2(1+2)+1/3(1+2+3)+.......+1/199(1+2+3+4+.....+199)
Tính: A = 1/2 + 1 + 1/2(1+2) + 1/3(1+2+3) + 1/4(1+2+3+4) + ... + 1/199(1+2+3+4+...+199)
Cho A=1/2+1/3+1/4+...+1/199+1/200
B=1/199+2/198+3/197+...+198/2+199/1
Tính A : B
\(B=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+....+\frac{198}{2}+\frac{199}{1}\)
\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+.....+\left(\frac{198}{2}+1\right)+\frac{200}{200}\)
\(=200\left(\frac{1}{100}+\frac{1}{199}+\frac{1}{198}+....+\frac{1}{2}\right)\)
= 200.A
=> A:B=\(\frac{1}{200}\)
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+.......+\frac{1+2+3+...+199}{199}=?\)= ....................
\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+3+...+199}{199}\)
\(\text{Đ}\text{ặt}S=1+\frac{1}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+199}{199}\)
\(\Rightarrow S=1+\frac{\left(2+1\right).2}{2}+\frac{\left(3+1\right)3}{3}+...+\frac{\left(199+1\right)199}{199}\)
\(S=1+\frac{2+1}{1}+\frac{3+1}{1}+...+\frac{199+1}{1}\)
\(\Rightarrow S=1+\left(3+4+...+200\right)\)
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
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\(1+\frac{1+2}{2}+\frac{1+2+3}{3}+...........+\frac{1+2+3......+199}{199}\)
Cho A= ½ + 1/3 + 2/4 +…+ 1/200 và B= 1/199+2/198+3/197+…+199/1
Ta có :
B = 1/ 199 + 2/ 198 + 3/197+...+ 1+ 1 + 1 + ....+ 1. ( tách 199/1 = tổng của 199 số 1)
B = 1 + ( 1+ 1/199) + (1 + 1/198) + ( 1+ 1/197) +....+ (1 + 198/2)
B = 200/200 + 200/199 + 200/198 + 200/197 +...+ 200/2
B = 200 x ( 1/200 + 1/199 + 1/198 + 1/197 +...+ 1/2)
=> A/B =1/ 200
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D=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)
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