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Ta có \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\) (đpcm)
Áp dụng công thức trên ta có
A\(=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\cdot\cdot\cdot\cdot\cdot\cdot\cdot+\frac{1}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{2015\cdot2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{2}{3\cdot4}+....+\frac{1}{2015\cdot2016}-\frac{1}{2016\cdot2017}\)
\(\Leftrightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\)
\(\Rightarrow A=\left(\frac{1}{1\cdot2}-\frac{1}{2016\cdot2017}\right)\div2\approx0.25\)
Vậy A\(\approx0.25\)
a) Nhân cả tử và mẫu với 2 . 4 . 6 ... 40 ta được :
\(\frac{1.3.5...39}{21.22.23...40}=\frac{\left(1.3.5...39\right).\left(2.4.6...40\right)}{\left(21.22.23...40\right).\left(2.4.6...40\right)}\)
\(=\frac{1.2.3...39.40}{1.2.3...40.2^{20}}=\frac{1}{2^{20}}\)
b) Nhân cả tử và mẫu với 2 . 4 . 6 ... 2n ta được :
\(\frac{1.3.5...\left(2n-1\right)}{\left(n+1\right)\left(n+2\right)\left(n+3....2n\right)}=\frac{1.3.5...\left(2n-1\right).\left(2.4.6...2n\right)}{\left(n+1\right)\left(n+2\right)...\left(2n\right).\left(2.4.6...2n\right)}\)
\(=\frac{1.2.3...\left(2n-1\right).2n}{1.2.3...2n.2^n}=\frac{1}{2^n}\)
Xét vế phải: \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
= \(\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
= \(\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}\)
= \(\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
= VT
=> Đpcm
\(A.\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}=\frac{1}{n.\left(n+1\right)}\left(ĐPCM\right)\)
\(B.\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n.\left(n+a\right)}-\frac{n}{n.\left(n+a\right)}=\frac{a}{n.\left(n+a\right)}\left(ĐPCM\right)\)
Tham khảo nha !!!!
a,
\(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
b,
\(\frac{a}{n\left(n+a\right)}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\)
CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)(1)
\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\)(2)
Từ (1) và (2) suy ra \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}.\frac{1}{n+1}\)(đpcm)
\(\frac{1}{n}-\frac{1}{\left(n+1\right)}\)
\(=\frac{n+1}{n.\left(n+1\right)}-\frac{n}{\left(n+1\right).n}\)
\(=\frac{n+1-n}{n.\left(n+1\right)}\)
\(=\frac{1}{n.\left(n+1\right)}\)
Ta có : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Vì VT=VP nên ta có đpcm
\(\text{Ta có:}\)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+1}+\frac{1}{n+2}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(1\right)\)
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}+\frac{2}{n+1}-\frac{2}{n+1}=\frac{2n\left(n+2\right)+2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)^2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(2\right)\)
\(\text{Từ (1) và (2) ta có: ĐPCM}\)
100 + 100 + 100
Các bạn trả lời nhanh nhất mình k cho mà bạn nào trả lời nhanh nhất thì các bạn k cho bạn đấy mình sẽ k lại cho
a) \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{q}\left(\frac{1}{n}-\frac{1}{n+q}\right)=\frac{1}{q}\left(\frac{n+q}{n\left(n+q\right)}-\frac{n}{n\left(n+q\right)}\right)=\frac{1}{q}.\frac{q}{n\left(n+q\right)}=\frac{1}{n\left(n+q\right)}\)
a/ Xét mẫu số VP_ n và n+1 là 2 số liên tiếp
\(\Rightarrow\left(n,n+1\right)\)bằng 1
Thay vào đề bài \(\frac{1}{n}-\frac{1}{n+1}\)bằng \(\frac{n+1}{n.\left(n+1\right)}-\frac{n}{n.\left(n+1\right)}\)bằng \(\frac{1}{n\cdot\left(n+1\right)}\)
\(\Rightarrowđpcm\)
P/s _laptop ko gõ đc dấu