Tính
B=1-\(\frac{1}{2}\)(1+2)-\(\frac{1}{3}\)(1+2+3)-\(\frac{1}{4}\)(1+2+3+4)-...-\(\frac{1}{20}\)(1+2+3+...+20)
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NN
\(\frac{2}{1+2}+\frac{2+3}{1+2+3}+\frac{2+3+4}{1+2+3+4}+......+\frac{2+3+4+...+20}{1+2+3+4+...+20}\)
0
4 tháng 7 2017
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
BT
16 tháng 5 2017
Ta có: 1+2+3+...+n=\(\frac{n\left(n+1\right)}{2}\)
=> \(1=\frac{1x2}{2};\frac{1}{2}\left(1+2\right)=\frac{2x3}{2x2};\frac{1}{3}\left(1+2+3\right)=\frac{3x4}{2x3};\)\(;\frac{1}{4}\left(1+2+3+4\right)=\frac{4x5}{2x4};...;\frac{1}{20}\left(1+2+3+...+20\right)=\frac{20x21}{2x20}\)
=> \(B=\frac{1x2}{2}+\frac{2x3}{2x2}+\frac{3x4}{2x3}+\frac{4x5}{2x4}+...+\frac{20x21}{2x20}\)
=> \(B=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
=> \(B=\frac{1}{2}\left(2+3+4+5+...+21\right)=\frac{1}{2}\left(\frac{21.22}{2}-1\right)\)
=> \(B=\frac{230}{2}=115\)
Đáp số: B=115
LT
0
22 tháng 1 2019
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+...+\frac{1}{20}.\left(1+...+20\right).\)
\(=1+\frac{3}{2}+\frac{6}{3}+\frac{10}{4}+...+\frac{210}{20}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=\frac{230}{2}=115\)
\(B=1-\frac{1}{2}\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3+4\right)-...-\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(B=1-\frac{1}{2}.\left(1+2\right).2:2-\frac{1}{4}.\left(1+4\right).4:2-...-\frac{1}{20}.\left(1+20\right).20:2\)
\(B=1-3:2-5:2-...-21:2\)
\(B=1-3.\frac{1}{2}-5.\frac{1}{2}-...-21.\frac{1}{2}\)
\(B=1-\frac{1}{2}.\left(3+5+...+21\right)\)
Đặt C = 3 + 5 + ... + 21
Số số hạng của tổng C là: (21 - 3) : 2 + 1 = 10 (số)
=> C = (3 + 21) x 10 : 2 = 24 x 5 = 120
=> \(A=1-\frac{1}{2}.120\)
\(A=1-60=-59\)