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1 Tính :
a) \(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{\left(n-1\right).n}\)
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{n}\)
\(=\frac{1}{n}\)
b) \(B=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(=\frac{4}{1.5}-\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{\left(n-4\right).n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\left(\frac{1}{5}-\frac{1}{n}\right)\)
\(=\frac{4}{5}-\frac{1}{5}+\frac{1}{n}\)
\(=\frac{3}{5}+\frac{1}{n}\)
c) \(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\)
\(=1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\Rightarrow C=1-B\left(1\right)\)
\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\)
Lấy 2B trừ B ta có :
\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(B=1-\frac{1}{2^{10}}\left(2\right)\)
Thay (2) vào (1) ta có :
\(C=1-\left(1-\frac{1}{10}\right)\)
\(=1-1+\frac{1}{10}\)
\(=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
a) S1 = \(-\frac{1}{1.2}-\frac{1}{2.3}-...-\frac{1}{99.100}\)
= \(-\frac{1}{1}-\frac{1}{2}-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{99}-\frac{1}{100}\)
= \(\frac{-1}{1}-\frac{1}{100}\)
= \(-\frac{101}{100}\)
M=\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right).n}\)
\(M=1-\frac{1}{5}-\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(M=1-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)
\(M=\frac{3}{5}+\frac{1}{n}\)
Mình chỉ giải đến đây thôi vì chẳng biết n bằng mấy cả
= - (1-1/5 +1/5 -1/9 +1/9 -1/13 +1/n + 1/n+4)
=-(1-1/n+4)
=-1+1/n+4
Có dạng tổng quát như thế này nhé:
\(\frac{k}{n\left(n+k\right)}=\frac{1}{n}-\frac{1}{k+n}\)
Trong trường hợp này là \(\frac{-4}{1.5}-\frac{4}{5.9}-...-\frac{4}{\left(n+4\right)n}=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)
Đáp án là: \(\frac{1}{n+4}-1\)
1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
a) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{1}-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Còn câu b niwax nha các bn . Giúp mk với