How to solve nonlinear systems of equation with summation I'm stuck with this
$$
\sum _{n=0}^{80}\:0.1+b^{n+k}=100
$$
$$
0.1+b^{80+k}=10
$$
Is it possible to solve?
 A: The summation of the first equation contains $81$ terms, from $n=0$ to $n=80$. Starting to count the terms from  the last one and calling them $a_0, a_1, a_2...a_{i}$ with $i=80-n$, we have  $a_0=10$.
Now we have $$a_1=(a_0-0.1)/b+0.1$$ and in general
$$a_{i+1}=(a_i-0.1)/b+0.1$$
This recurrence has solution
$$a_i = \frac{99/b^i + 1}{10} $$
So we can write
$$\sum_{i=0}^{80} \frac{99/ b^{i} + 1}{10}=100 $$
$$\sum_{i=0}^{80} \left(\frac{99}{ b^{i}} + 1\right)=1000 $$
$$\sum_{i=0}^{80} \frac{99}{b^{i}}=1000-81=919 $$
$$\sum_{i=0}^{80} \frac{1}{ b^{i}}=\frac{919}{99} $$
$$\frac{b^{81}-1}{ b^{80}(b-1)}=\frac{919}{99} (\text{with 
  } b\neq 1)$$
$$b^{81}-1 =\frac{919}{99} b^{80}(b-1)$$
$$\frac{820}{99} b^{81} - \frac{919}{99} b^{80}+1=0$$
$$820 b^{81} - 919 b^{80}+99=0$$
This last equation has three real solutions, that WA numerically determines here :
$$b=1$$
$$b=-0.96501...$$
$$b=1.12071...$$
The first solution cannot be accepted because of the above condition $b\neq 1$ (also note that $b=1$ trivially could not satisfy the initial equations of the OP).
Because from the second equation of the OP we get $$k=\log(9.9)/\log(b)-80$$
only the third solution is acceptable, providing
$$k\approx \frac{\log(9.9)}{\log(1.12071)}-80 \approx -59.884...$$
Here WA computes the summation in the LHS of the first equation of the OP using the values of $b$ and $k$ reported above, giving confirmation of the expected result.
