Is mathematical Induction possible in this situation? Is mathematical Induction possible with this sigma sign? 
$\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
with $n = 2s+1 ;  s \epsilon \mathbb{N}$
Statement: $\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
Assumption: $\sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n+2}+1}{b+1}$
I already checked the basis but I have problems with splitting up the assumption.
I tried it successfully with:
$\sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) \equiv  \sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) + (-1)^{n+1}*b^{n+1} + (-1)^{n}*b^{n}$
$\Leftrightarrow \sum_{k=1}^{n+2} ((-1)^{n-k} * b^{n-k}) \equiv  \sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) + b^{n+1} - b^{n}$
It is kind of consequentially if you look at this:
$\frac{b^{n}+1}{b+1}$
$\frac{b^{3}+1}{b+1} = b^{2}-b+1$
$\frac{b^{5}+1}{b+1} = b^{4} - b^{3} + b^{2}-b+1$
$
\frac{b^{7}+1}{b+1} = b^{6}-b^{5}+ b^{4} - b^{3} + b^{2}-b+1$
but I cant find a mathematical correct way to split my sigma sign like above.
I am pretty sure that the Statement is correct for any possible natural uneven number. 
Now Im not even sure if i used the right Assumption for n -> n+2.
I would be very happy with some help :)
 A: First, you need the statement you are trying to prove to be correct.  You want your sums to go up to an even number, not to an odd number.  If you look at your examples the highest power on the right is even.  The power of $b$ on the right should be one greater than the upper limit of the sum.  You also start the sum with $1=b^0$, so the lower limit of the sum should be $k=0$. So you are trying to prove $$\sum_{k=0}^{n-1} ((-1)^{n-k-1} * b^{n-k}) = \frac{b^{n}+1}{b+1}$$ with $n$ odd.  What you call the Statement is the induction hypothesis.  You need a base case, presumably $n=3$, which you show at the bottom.  Then what you call Assumption is the thing you are trying to prove from the induction hypothesis.  Note that there are just two extra terms added to the prior sum, which is how you make use of the induction hypothesis.
$$\begin {align} \sum_{k=0}^{n+1} ((-1)^{n-k-1} * b^{n-k}) &=b^{n+1} - b^{n}+\sum_{k=0}^{n-1} ((-1)^{n-k} * b^{n-k-1})\\&=b^{n+1}-b^{n}+\frac {b^n+1}{b+1}\\&=\frac{b^{n+2}+1}{b+1}\end {align}$$
A: Let's pose
$$A_s = \sum_{k=1}^{2s+1} (-b)^{2s+1-k} = \frac{b^{2s+1}+1}{b+1}.$$
We start from $A_1$:
$$A_1 = b^2-b+1 = \frac{b^3+1}{b+1} = \frac{b^{2s+1}+1}{b+1}.$$
Now we need to find the inductive rule.
$$A_{s+1} = \sum_{k=1}^{2(s+1)+1} (-b)^{2(s+1)+1-k} = \sum_{k=1}^{2s+3} (-b)^{2s+3-k} = \\
= \sum_{k=1}^{2s+1} (-b)^{2s+3-k} + (-b)^{2s+3-(2s+2)} + (-b)^{2s+3-(2s+3)}= \\
= (-b)^2\sum_{k=1}^{2s+1} (-b)^{2s+1-k} + (-b)^{1} + (-b)^{0}= \\
= b^2 A_s -b + 1.$$
Finally:
$$A_{s+1} = b^2 A_s -b + 1 = b^2 \frac{b^{2s+1}+1}{b+1} -b + 1 = \frac{b^{2s+3}+b^2+(-b+1)(b+1)}{b+1} = \\
= \frac{b^{2(s+1)+1}+b^2-b^2 + 1}{b+1} = \frac{b^{2(s+1)+1} + 1}{b+1}.$$
