Ta có
$$\dfrac{321^{22}}{320.32^{22}} > \dfrac{320^{22}}{320.32^{22}}$$
$$= \dfrac{(32.10)^{22}}{320.32^{22}}$$
$$= \dfrac{32^{22}.10^{22}}{320.32^{22}}$$
$$= \dfrac{10^{22}}{320}$$
$$= \dfrac{10^{21}}{32}> \dfrac{100}{32} > 1$$
Vậy ta có
$$\dfrac{321^{22}}{320.32^{22}} >1$$
hay $321^{22} > 320.32^{22}$.
b) Xét tổng
$$A = 1 + 2002 + 2002^2 + \cdots + 2002^{99}$$
Ta có
$$2002A = 2002 + 2002^2 + 2002^3 + \cdots + 2002^{99} + 2002^{100}$$
Khi đó
$$2002A - A = 2002 + 2002^2 + 2002^3 + \cdots + 2002^{99} + 2002^{100} - (1 + 2002 + 2002^2 + \cdots + 2002^{99}) = 2002^{100}-1$$
Vậy
$$A = \dfrac{2002^{100}-1}{2001} < 2002^{100}$$
Vậy
$$1 + 2002 + 2002^2 + \cdots + 2002^{99} < 2002^{100}$$
