Prove by induction that the sum of the terms in a sequence is: $\frac{a_0 - a_{n+1} }{1 - r} $

Question

I am trying to prove the following by induction.

Let the follownig finite sequence $ a_0, a_1, a_2, a_3,....,a_n $ be defined by the rule: $$ a_i = a_0 \cdot r^i$$ where r is some number not equal to 1.

Prove by induction that the sum of the terms in the sequence is $$ \frac{a_0-a_{n+1}}{1 - r}$$ $ a_{n+1} $ is defined by the sequence, even though it is not part of the sequence.

I am completely lost on how to approached this problem.

Answer 1

Note that for $n=1$ we have $$ a_{0} + a_{1} = a_{0}(1+r) = a_{0}\frac{1-r^{2}}{1-r} = \frac{a_{0} - a_{o}r^{2}}{1-r} = \frac{a_{0} - a_{2}}{1-r}. $$ If $n \geq 1$ is an integer such that $$ \sum_{i=1}^{n-1}{a_{i}} = \frac{a_{0} - a_{n}}{1-r}, $$ then $$ \sum_{i=1}^{n}a_{i} = \frac{a_{0}-a_{n}}{1-r} + a_{n} = \frac{a_{0}-a_{n} + a_{n} - ra_{n}}{1-r} = \frac{a_{0} - ra_{n}}{1-r} = \frac{a_{0} - a_{n+1}}{1-r}. $$

Answer 2

For $n+1$ you have:

$$a_0+a_1+...+a_n+a_{n+1}=\frac{a_0-a_{n+1}}{1-r}+a_{n+1}=\\ \frac{a_0-a_{n+1}+(1-r)a_{n+1}}{1-r}=\frac{a_0-ra_{n+1}}{1-r}=\frac{a_0-a_{n+2}}{1-r}$$

For the last equality we used that

$$ra_{n+1}=ra_0r^{n+1}=a_0r^{n+2}=a_{n+2}$$

Answer 3

Let ${ S }_{ n }=\frac { a_{ 0 }-a_{ n+1 } }{ 1-r } $ we should show that $${ S }_{ n+1 }=\frac { a_{ 0 }-a_{ n+2 } }{ 1-r } $$ $$\underset { { S }_{ n } }{ \underbrace { { a }_{ 0 }+{ a }_{ 1 }+...+{ a }_{ n } } } +{ a }_{ n+1 }=\frac { a_{ 0 }-a_{ n+1 } }{ 1-r } +{ a }_{ n+1 }=\frac { a_{ 0 }-a_{ n+1 }+{ a }_{ n+1 }-r{ a }_{ n+1 } }{ 1-r } =\frac { { a }_{ 0 }-r{ a }_{ n+1 } }{ 1-r } =\frac { { a }_{ 0 }-{ a }_{ n+2 } }{ 1-r } $$