Question

[Thị Huyền Trang Nguyễn]

3, Chứng tỏ :

\(B=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}< 1\)

Answer 1

[Đỗ Thanh Hải]

Ta có

B = \(\dfrac{1}{2!}\) + \(\dfrac{2}{3!}\) + \(\dfrac{3}{4!}\) + ..... + \(\dfrac{99}{100!}\)

B = \(\dfrac{2-1}{2!}\) + \(\dfrac{3-1}{3!}\) + \(\dfrac{4-1}{4!}\) + ... + \(\dfrac{100-1}{100!}\)

B = \(\dfrac{2}{2!}\) - \(\dfrac{1}{2!}\) + \(\dfrac{3}{3!}\) - \(\dfrac{1}{3!}\) + ... + \(\dfrac{100}{100!}\) - \(\dfrac{1}{100!}\)

B = 1 - \(\dfrac{1}{2!}\) + \(\dfrac{1}{2!}\) - \(\dfrac{1}{3!}\) + ... + \(\dfrac{1}{99!}\)- \(\dfrac{1}{100!}\)

B = 1 - \(\dfrac{1}{100!}\) < 1

=> B < 1 <đpcm>

Answer 2

[Trần Huỳnh Tuyết Trinh]

B=\(\dfrac{1}{2!}\)+\(\dfrac{2}{3!}+\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\)

=\(\dfrac{2-1}{2!}\)+\(\dfrac{3-1}{3!}+\dfrac{4-1}{4!}\)+...+\(\dfrac{100-1}{100!}\)

=\(\dfrac{2}{2!}-\dfrac{1}{2!}+\dfrac{3}{3!}-\dfrac{1}{3!}+\dfrac{4}{4!}-\dfrac{1}{4!}+...+\dfrac{100}{100!}-\dfrac{1}{100!}\)

=\(\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\dfrac{1}{99!}-\dfrac{1}{100!}\)

=\(1-\dfrac{1}{100!}\)< 1

\(\Rightarrow\)B =\(\dfrac{1}{2!}\)+\(\dfrac{2}{3!}+\dfrac{3}{4!}\)+...+\(\dfrac{99}{100!}\) < 1

Chúc bạn học tốt !