How do you sum the series #1+a+a^2+a^3+...+a^n#?

2 Answers
Jul 3, 2015

Answer:

#1+a+a^2+a^3+...+a^n = (1-a^(n+1))/(1-a)#

Explanation:

#(1+a+a^2+a^3+...+a^n)(1-a)=1-a^(n+1)#

Divide both sides by #(1-a)# to get:

#1+a+a^2+a^3+...+a^n = (1-a^(n+1))/(1-a)#

Jul 3, 2015

Answer:

The solution George C.gives is very elegant. Here's another that has its own uses.

Explanation:

Let

#S_n = 1 + a + a^2 +a^3 + * * * +a^n#

Multiply by #a# to get:

#aS_n = " " a + a^2 +a^3 + * * * +a^n +a^(n+1)#

Now subtract:

#(1-a)S_n = 1-a^(n+1)#

So, again:

#S_n = (1-a^(n+1))/(1-a)#