Question

Chứng minh rằng : \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+......+\dfrac{99.100-1}{100!}< 2\)

GIÚP MÌNH VỚI

Answer 1

Ta có:

\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}\)

\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+...+\dfrac{99.100}{100!}-\dfrac{1}{100!}\)

\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+...+\dfrac{99.100}{100!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)

\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)

\(=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}\)

\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)

Vậy \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}< 2\) (Đpcm)