Question

4+3^1+3^2+3^3+...+3^100=?

Answer 1

Đặt A= 4+3^1+3^2+3^3+...+3^100

Đặt B = 3+3²+3³+3⁴+......+3¹⁰⁰

3B= 3²+3³+3⁴+......+3¹⁰⁰⁺3¹⁰¹

3B-B= (3²+3³+3⁴+......+3¹⁰⁰⁺3¹⁰¹)-(3+3²+3³+3⁴+......+3¹⁰⁰)

2B= 3¹⁰¹-3

B=  (3¹⁰¹-3) : 2

=>A=4+(3¹⁰¹-3) : 2

Answer 2

Ta gọi tổng đó là A

Ta có:

A=3+3²+3³+...+3¹⁰⁰+4

3A=3²+3³+...+3¹⁰¹+12

3A-A=3¹⁰¹-3-8

A=(3¹⁰¹-3-8):2