đặt tổng này là S
ta có:
3S=3[1.2+2.3+...+(n-2)(n-1)+(n-1)n]
3S=1.2.3+2.3.3+...+(n-2)(n-1).3+(n-1)n.3
3S=1.2.(3-0)+2.3.(4-1)+...+(n-2)(n-1)[(n+3)-n]+(n-1).n.[(n-1)+(4-n)]
3S=1.2.3+2.3.4-1.2.3+...+(n-2)(n-1)(n+3)-(n-2)(n-1)n+n(n-1)+n(4-n)
3S=n(n-1)[(n-2)(n-1)(n+3)+n(4-n)]
S=n(n-1)[(n-2)(n-1)(n+3)+n(4-n)]:3
\(A=1.2+2.3+...+n\left(n+1\right)\)
\(\Rightarrow3A=1.2.3+2.3.3+...+n\left(n+1\right)3\)
\(=1.2.3+2.3.\left(4-1\right)+...+n\left(n+1\right)\left[\left(n+2\right)-\left(n-1\right)\right]\)
\(=1.2.3+2.3.4-1.2.3+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right)n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Tính giá trị \(C=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)^2\right]}\)
\(\frac{-1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{\left(n-1\right).n}\left(n\in N\ne0,n\ne1\right)\)
Chứng minh rằng với mọi n \(\inℕ^∗\):
D = \(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}< 1\)
F = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
\(\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)\left(\frac{2}{4.5}\right)........\left(\frac{2}{n\left(n+1\right)}-1\right)\left(n\in N\ne0,n\ge2\right)\)
CMR : 1.2+2.3+3.4+...+n.(n+1)=\(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
CM bằng phương pháp quy nạp toán học nha
nhớ quy nạp
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
tính
A = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(3.2\right)^2}+.....+\frac{2n+1}{\left[n.\left(n+1\right)\right]^2}\)
Tính giá trị của biểu thức A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
