Ta có :
\(A=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+...+\frac{\left(n-1\right)n-1}{n!}\)
\(=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+\frac{3.4}{4!}-\frac{1}{4!}+...+\frac{\left(n-1\right)n}{n!}-\frac{1}{n!}\)
\(=1-\frac{1}{2!}+1-\frac{1}{3!}+\frac{1}{2!}-\frac{1}{4}!+\frac{1}{3!}-\frac{1}{5!}+\frac{1}{4!}-...+\frac{1}{\left(n-2\right)!}-\frac{1}{n!}\)
\(=2-\frac{1}{n!}< 2\)
Vậy ...
Bài 1 :Chứng tỏ rằng
D=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+...+\frac{2!}{n!}< 1\)
Bài 2 :Chứng minh rằng \(\forall n\in Z\left(n\ne0,n\ne1\right)\)thì \(Q=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\)không phải số nguyên
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 = \(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{n\left(n+2\right)}\right)< 2\)
\(F=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=\frac{n-1}{n}\)
\(G=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)\left(n-1\right).n}=\)
\(H=2+4+6+..+2n=\)
1.Tính:
a,\(A=\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-......-\frac{1}{\left(n-1\right).n}\)\(n\in N\)
b,\(\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-....-\frac{4}{\left(n-4\right).n}\)\(n\in N\)
c\(C=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-.....-\frac{1}{2^{10}}\)
\(\frac{-1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-....-\frac{1}{\left(n-1\right).n}\left(n\in N\ne0,n\ne1\right)\)
\(\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)\left(\frac{2}{4.5}\right)........\left(\frac{2}{n\left(n+1\right)}-1\right)\left(n\in N\ne0,n\ge2\right)\)
Cho n!=1.2.3....n,đọc là n giai thừa.Chứng minh rằng:
a.\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+....+\frac{99}{100!}< \)\(1\)
b.\(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+....+\frac{99.100-1}{100!}< 2\)
Tính tổng
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{n\left(n+1\right)}\)
\(\text{Chứng minh rằng : }\)\(\forall n\in Z\left(n\ne0,n\ne-1\right)\)\(\text{thì }\)\(Q=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n\left(n+1\right)}\)
\(\text{Không phải là số nguyên}\)
