<s>[INST] Prove by induction that $4\sum_{k=-1}^{n}(5)^k = 5^{n+1}-\frac{1}{5}, \forall n \in \mathbb{N}$ I have the following:
using the principle of mathematical induction, prove that is true that $\forall n \in \mathbb{N}$ we have
$$4\sum_{k=-1}^{n}(5)^k = 5^{n+1}-\frac{1}{5}$$
I have done it, but with some problems.
I consider the statement:
$$P(n) :4\sum_{k=-1}^{n}(5)^k = 5^{n+1}-\frac{1}{5}, \quad \forall n \in \mathbb{N}$$
since $n \in \mathbb{N}$, it is a number $\ge0$, so the base step is to value the statement for $n=0$.
$$\begin{array}{lcl}P(0) : 4\sum_{k=-1}^{0}(5)^k & = & 5^{0+1}-\frac{1}{5} \end{array}$$ hence, $$\begin{array}{lcl}4 \left ( 5^{-1} + 5^0\right ) & = & 5 -\frac{1}{5} \\ \frac{24}{5} &= & \frac{24}{5} \end{array}$$
so it is true for $n=0$.  
Now assume as true the statement for $n=h$
$$P(h) : 4\sum_{k=-1}^{h}(5)^k = 5^{h+1}-\frac{1}{5}$$
basing on the truth of this last one, let's check the truth of $P(h+1)$:
$$P(h+1) : 4\sum_{k=-1}^{h+1}(5)^k = 5^{(h+1)+1}-\frac{1}{5}$$
now here I encounter some problems, I know that inside this last summation is included both $h$ and $h+1$ but I don't understand how to do, I have tried this without success:
$$\begin{array}{rcl}4\sum_{k=-1}^{h+1}(5)^k & = & 4\sum_{k=-1}^{h}(5)^k + \sum_{k=h+1}^{h+1}(5)^k \\ & = & 5^{h+1} -\frac{1}{5} + 5^{h+1} \\ & = & 2 \cdot 5^{h+1} -\frac{1}{5} \end{array}$$
it not the result $5^{(h+1)+1} - \frac{1}{5}$ desiderd.
Please, can you tell me, Where is the error? Thanks!
 [/INST] The problem is in the last line. We have 
$\begin{array}{rcl}4\sum_{k=-1}^{h+1}(5)^k & = & 4\sum_{k=-1}^{h}(5)^k + 4\sum_{k=h+1}^{h+1}(5)^k \\ & = & 5^{h+1} -\frac{1}{5} + 4 \cdot 5^{h+1} \\ & = & 5 \cdot 5^{h+1} -\frac{1}{5} \\
& = & 5^{h+2} - \frac{1}{5}\end{array}$
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