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Đặt C =\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow2C=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow C=\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\div2\)
Đặt tổng trên là A
Có : 3A = 1.2.3+2.3.3+....+n.(n+1).3
= 1.2.3+2.3.(4-1)+......+n.(n+1).[(n+2)-(n-1)]
= 1.2.3+2.3.4-1.2.3+.....+n.(n+1).(n+2)-(n-1).n.(n+1)
= n.(n+1).(n+2)
=> A = n.(n+1).(n+2)/3
Tk mk nha
Đặt A=1.2+2.3+...+n(n+1)
3A=1.2.3+2.3.3+...+n(n+1).3
3A=1.2.(3-0)+2.3.(4-1)+...+n(n+1)[(n+2)-(n-1)]
3A=1.2.3-0.1.2+2.3.4-1.2.3+...+n(n+1)(n+2)-(n-1)n(n+1)
3A=[1.2.3+2.3.4+...+n(n+1)(n+2)]-[0.1.2+1.2.3+...+(n-1)n(n+1)]
3A=n(n+1)(n+2)-0.1.2
3A=n(n+1)(n+2)
A=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Gọi A=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
4A=1.2.3+2.3.4+3.4.5+...+n(n+1)(n+2)
=> 4A=1.2.3(4-0)+2.3.4(5-1)+...+n(n+1)(n+2)[(n+3)-(n-1)]
=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+n(n+1)(n+2)(n+3)-(n-1).n(n+1)(n+2)
=n(n+1)(n+2)(n+3)
4A+1=n(n+1)(n+2)(n+3)+1=n4+6.n3+11.n2+6n+1=(n2+3n+1)2
=>\(\sqrt{4A+1}\)=n2+3n+1
chúng ta hãy quy đồng rồi cộng chúng lại với nhau thì sẽ ra kết quả và cậu hãy xem lai kiến thức mới học của cậu đi
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(\Rightarrow F=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{\left(n-1\right)}-\frac{1}{n}\)
\(\Rightarrow F=1-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\left(đpcm\right)\)
\(H=2+4+6+...+2n\)
\(P=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(P=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(P=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(P=\left(x^2+5x\right)^2\ge-36\)
\(\Rightarrow GTNN\) của \(P=-36\)
Dấu = sảy ra khi:\(x^2+5x=0\)
.....................\(\Rightarrow x=0\) hoặc \(x=-5\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+....+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+2\right)+1\left(n+2\right)}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+2n+n+2}\right)\)
\(S_n=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2}\right)\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2\left(n^2+3n+2\right)}\)
\(S_n=\dfrac{1}{4}-\dfrac{1}{2n^2+6n+4}\)
\(S_n=\dfrac{2n^2+6n+4}{4\left(2n^2+6n+4\right)}-\dfrac{4}{4\left(2n^2+6n+4\right)}\)
\(S_n=\dfrac{2n^2+6n+4}{8n^2+48n+16}-\dfrac{4}{8n^2+48n+16}\)
\(S_n=\dfrac{2n^2+6n}{8n^2+48n+16}\)
\(S_n=\dfrac{2\left(n^2+3n\right)}{2\left(4n^2+24n+8\right)}=\dfrac{n^2+3n}{4n^2+24n+8}\)
\(S_n=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\\ 2S_n=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\\ =\dfrac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\\ =>S_n=\dfrac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)
Giải sai r nhéLinh Nguyễn
Đặt A là tên biểu thức
A=1.2.3+2.3.4+...+n(n+1)(n+2)
4A=1.2.3.4+2.3.4.4+...+n(n+1)(n+2).4
4A=1.2.3.4 - 0.1.2.3 + 2.3.4.5 - 1.2.3.4 +...+ n(n+1)(n+2)(n+3) - (n-1)n(n+1)(n+2)
4A=[1.2.3.4+2.3.4.5+...+n(n+1)(n+2)(n+3)] - [0.1.2.3+1.2.3.4+...+(n-1)n(n+1)(n+2)]
4A=n(n+1)(n+2)(n+3)-0.1.2.3
A=\(\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)
\(A=1.2.3+2.3.4+3.4.5+...+n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.4+3.4.5.4+...+4n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow4A=1.2.3.4+1.2.3.\left(5-1\right)+...+n\left(n+1\right)\left(n+2\right)\left(n+3-n+1\right)\)
\(\Rightarrow4A=1.2.3.4+2.3.4.5-1.2.3.4+...+n\left(n+1\right)\left(n+2\right)\left(n+3\right)-\left(n-1\right)n\left(n+1\right)\left(n\right)\)
\(\Rightarrow4A=n\left(n+1\right)\left(n+2\right)\left(n+3\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{4}\)