Tính tổng:
\(1.2+2.3+3.4+...+n\left(n+1\right)\)
\(1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(\Rightarrow F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(\Rightarrow F=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\left(đpcm\right)\)
\(H=2+4+6+...+2n\)
Tính các tổng :
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\) ( Hướng dẫn : \(\frac{1}{k\left(k+1\right)}=\frac{1}{k}-\frac{1}{k+1}\))
b) \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
( Hướng dẫn : \(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+2}\right)-\frac{1}{k+1}\))
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
a) Ta có: \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\)
\(A=1-\frac{1}{n+1}\)
\(A=\frac{n+1}{n+1}-\frac{1}{n+1}\)
\(A=\frac{n}{n+1}\)
Học tốt nha^^
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2B=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\)
\(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)
\(=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
??? Cái gì đây, đây là câu hỏi hay câu trả lời ???
Bài làm mà mấy thánh cứ vào phốt thế
Tính tổng : 1.2 + 2.3 + 3.4 + …..+ n.(n+1)
1.2.3+ 2.3.4 + 3.4.5 + ….+ n(n+1)(n+2)
https://olm.vn/hoi-dap/tim-kiem?q=t%C3%ADnh+t%E1%BB%95ng+sau+:S+=+1.2.3+2.3.4+3.4.5+...+n.(n+1).(n+2)+&id=601088
Sn = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
Tính tổng
Lời giải: Sử dụng hằng đẳng thức \(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\) ta có:
Sn=\(\frac{1}{2}\left[\frac{1}{1\times2}-\frac{1}{2\times3}\right]+\frac{1}{2}\left[\frac{1}{2\times3}-\frac{1}{3\times4}\right]+...\)\(+\frac{1}{2}\left[\frac{1}{\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
\(=\frac{1}{2}\left[\frac{1}{1\times2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]=\frac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right).\left(n+2\right)}\)
help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)=\(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
- Với \(n=1\Rightarrow1.2.3=\frac{1.2.3.4}{4}\) (đúng)
- Giả sử biểu thức đúng với \(n=k\) hay:
\(1.2.3+...+k\left(k+1\right)\left(k+2\right)=\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}\)
Ta cần chứng minh nó đúng với \(n=k+1\) hay:
\(1.2.3+...+k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)\left(k+3\right)=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}{4}\)
Thật vậy, ta có:
\(1.2.3+...+k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)\left(k+3\right)\)
\(=\frac{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}{4}+\left(k+1\right)\left(k+2\right)\left(k+3\right)\)
\(=\left(k+1\right)\left(k+2\right)\left(k+3\right)\left[\frac{k}{4}+1\right]\)
\(=\left(k+1\right)\left(k+2\right)\left(k+3\right).\frac{\left(k+4\right)}{4}\)
\(=\frac{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}{4}\) (đpcm)
Tìm x biết
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2013.2014.2015}\right)x=\left(1.2+2.3+3.4+.....+2014.2015\right)\)
\(\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{2013.2014}-\frac{1}{2014.2015}\right)x=\frac{1}{3}\left(2014.2015.2016-2013.2014.2015........+2.3.4-1.2.3+1.2.3-0.1.2\right)\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
\(x=\frac{1}{3.2029104}.2014^2.2015^2.2016=\)
\(\left(\frac{1}{2}-\frac{1}{2014.2015}\right)x=\frac{1}{3}.2014.2015.2016\)
Tính tổng \(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+......+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
Vậy..
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)=\dfrac{n^2+3n+2-2}{4\left(n+1\right)\left(n+2\right)}=\dfrac{n\left(n+3\right)}{4\left(n+1\right)\left(n+2\right)}\)
2B=\(\dfrac{2}{1.2.3}\)+\(\dfrac{2}{2.3.4}\)+....+\(\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{1.2}\)-\(\dfrac{1}{2.3}\)+\(\dfrac{1}{2.3}\)-\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{n\left(n+1\right)}\)-\(\dfrac{1}{\left(n+1\right)\left(n+2\right)}\)
=\(\dfrac{1}{2}\)-\(\dfrac{1}{n^2+3n+2}\)
=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\)
=>B=\(\dfrac{n^2+3n}{2\left(n^2+3n+2\right)}\):2
đến đây bạn tự tính nha !