Question

A = 3^1+3^2+...+3^100 tìm chữ số tận cùng

Answer 1

Ta có ; A = 3 + 3² + ..... + 3¹⁰⁰

=> 3A = 3² + 3³ + ..... + 3¹⁰¹ 

=> 3A - A = 3¹⁰¹ - 3

=> 2A = 3¹⁰¹ - 3

=> A = \(\frac{3^{101}-3}{2}\)

=> A = \(\frac{3^{100}.3-3}{2}=\frac{\left(3^{20}\right)^5.3-3}{2}=\frac{\left(....01\right)^5.5-3}{2}=\frac{\left(....01\right).5-3}{2}=\frac{\left(......05\right)-3}{2}\)

=> A = \(\frac{\left(....2\right)}{2}=\left(....1\right)\)

Answer 2

a﴿ 3S=3^2+3^3+3^4+...+3^101

=>3S‐S=﴾3^2+3^3+3^4+..+3^101﴿‐﴾3+3^2+3^3+...+3^100﴿

=>2S=3^101‐3 =>2S+3=3^101‐3+3=3^101

=>đpcm 

Vậy kp bài này là 1