Question

Tính tổng

S= 1+a+a mũ 2 + ... +a mũ n

giúp mình vs ạ

 

Answer 1

\(S\cdot a=a+a^2+...+a^{n+1}\\ S\cdot a-S=a+a^2+...+a^{n+1}-1-a-...-a^n\\ S\left(a-1\right)=a^{n+1}-1\\ S=\dfrac{a^{n+1}-1}{a-1}\)

Answer 2

\(S=1+a+a^2+...+a^n\)

\(\Rightarrow a.S=a+a^2+a^3+...+a^{n+1}\)

\(\Rightarrow\left(a-1\right)S=a.S-S=a+a^2+a^3+...+a^{n+1}-1-a-a^2-...-a^n=a^{n+1}-1\)

\(\Rightarrow S=\dfrac{a^{n+1}-1}{a-1}\)