## Task 24/64

Provide a summation formula for the series

$$
\sum_{\nu=0}^{n} \nu b \cdot e^{\nu(a+b x)}
$$
For $a \neq -x$ (we will consider this case first)

$$
\sum_{\nu=0}^{n} \nu b e^{\nu(a+b x)}=0 \cdot b e^{0}+b e^{a+b x}+2 b e^{2(a+b x)}+\ldots+n b e^{n(a+b x)}
$$

and since $\nu b e^{\nu(a+b x)}=\left(e^{\nu(a+b x)}\right)^{\prime}$, we can write

$$
\sum_{\nu=0}^{n} \nu b e^{\nu(a+b x)}=\sum_{\nu=0}^{n}\left[b e^{\nu(a+b x)}\right]^{\prime}=\left(e^{0}\right)^{\prime}+\left(e^{a+b x}\right)^{\prime}+\ldots+\left(e^{n(a+b x)}\right)^{\prime}=\left[e^{0}+e^{a+b x}+\ldots+e^{n(a+b x)}\right]^{\prime}
$$

Inside the square brackets, we have a geometric series with $a_{1}=e^{0}=1 ; q=e^{a+b x}$ and with $n+1$ terms, whose sum can be expressed as

$$
s_{n+1}=a_{1} \frac{q^{n+1}-1}{q-1}=\frac{e^{(n+1)(a+b x)}-1}{e^{a+b x}-1}
$$

Thus, we have

$$
\sum_{\nu=0}^{n} \nu b e^{\nu(a+b x)}=\sum_{\nu=0}^{n}\left[b e^{\nu(a+b x)}\right]^{\prime}=\left[\frac{e^{(n+1)(a+b x)}-1}{e^{a+b x}-1}\right]^{\prime}=\frac{b\left[n e^{(n+2)(a+b x)}-(n-1) e^{(n+1)(a+b x)}+e^{a+b x}\right]}{\left(e^{a+b x}-1\right)^{2}}
$$

(according to the quotient rule of differential calculus).

For the case $a=-b x$, we obtain an arithmetic series:

$$
\sum_{\nu=0}^{n} \nu b e^{\nu(a+b x)}=\sum_{\nu=0}^{n} \nu b=b \frac{n(n+1)}{2}
$$