<s>[INST] 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 :)
 [/INST] 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:
$$A_{s+1} = b^2 A_s -b + 1 = b^2 \frac{b^{2s+1}+1}{b+1} -b + 1 = \frac{b^{2s+3}+b^2+(-b+1)(b+1)}{b+1} = \\
= \frac{b^{2(s+1)+1}+b^2-b^2 + 1}{b+1} = \frac{b^{2(s+1)+1} + 1}{b+1}.$$
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