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\(=-\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{61}-\dfrac{1}{64}\right)=-\dfrac{1}{63}\)
Làm từng phần nha bạn
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)
Đặt \(A+x=\frac{299}{301}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)
\(A=1-\frac{1}{301}\)
\(A=\frac{300}{301}\)
=> \(\frac{300}{301}+x=\frac{299}{301}\)
\(x=\frac{299-300}{301}\)
\(x=-\frac{1}{301}\)
\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)
\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=\frac{303}{304}\)
\(A=\frac{505}{304}\)
dễ mà
Gọi tổng đó là S. Theo đề \(S=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{40.43}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{40}-\frac{1}{43}\)
\(S=1-\frac{1}{43}=\frac{42}{43}\)
3/1.4+3/4.7+3/7.10+3/10.13
=1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13
=1 - 1/13
=12/13
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
Chúc bạn học tốt
A= 3/1.4+ 3/4.7+ 3/7.10+...+ 3/27.30
= 1/3 . ( 3- 3/4 + 3/4- 3/7 +3/7- 3/10+... + 3/27- 3/30)
= 1/3 . (3 - 3/30)= 1/3 . 87/30= 87/90=29/30
A = 1/1 -1/4 +1/4 - 1/7 +1/7 ........+1/40 - 1/43
A = 1/1 - 1/43
A = 42/43
A=1 - 1/4 + 1/4 - 1/7 + .... + 1/40 - 1/43
= 1 - 1/43
= 42/43
3/1.4 + 3/4.7 + .. +3/13.16
= 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16
= 1/1 - 1/16
= 15/16
= 1-1/4+1/4-1/7+1/7-1/10 +...+ 1/x -1/ x+3
= 1 -1/x+3
= x+2 / x+3
Câu này dễ mà.
Sửa đề: \(M=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2022\cdot2025}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{2025}\)
\(=1-\dfrac{1}{2025}=\dfrac{2024}{2025}\)