Question

Cho T= 2019⁰+ 2019¹+2019²+...+2019²⁰¹¹

Tính 2018T+1

Answer 1

\(T=2019^0+2019^1+2019^2+...+2019^{2011}\)

\(\rightarrow2019T=2019\left(2019^0+2019^1+2019^2+...+2019^{2011}\right)\)

\(\rightarrow2019T=2019^1+2019^2+2019^3+...+2019^{2012}\)

\(\rightarrow2019T-T=(2019^1+2019^2+2019^3+...+2019^{2012})-\left(2019^0+2019^1+...+2019^{2011}\right)\)

\(\rightarrow2018T=2019^{2012}-2019^0=2019^{2012}-1\)

\(\rightarrow2018T+1=2019^{2012}-1+1=2019^{2012}\)

Answer 2

T = 2019⁰ + 2019¹ + 2019² +...+2019²⁰¹¹

T = 1 + 2019¹ + 2019² +...+ 2019²⁰¹¹

2019T = 2019¹ + 2019² +2019³ +...+2019²⁰¹²

2019T - T = (2019¹ + 2019² +2019³ +...+2019²⁰¹²) - (1 + 2019¹ + 2019² +...+ 2019²⁰¹¹)

2018T = 2019²⁰¹² - 1

=> 2018T + 1 = 2019²⁰¹²