I'm looking for a function that generates the sum of the products of the $i$-terms of a geometric and arithmetic progressions. In other words, I'm searching a closed form for the expression: $$\Omega_n=\tau_0+\tau_1+\tau_2+\cdots+\tau_n=\sum_{i=0}^{n-1}\tau_i=\sum_{i=0}^{n-1}\phi_i\cdot\Phi_i$$
where $\phi$ is an arithmetic progression: $$\phi_i = \mu_{\phi}+\delta_{\phi}\cdot i$$ and $\Phi$ is a geometric progression: $$\Phi_i=\mu_{\Phi}\cdot\delta_{\Phi}^i$$
So far, I've done those steps:
-Rewrite the summations as: $$\Omega_n=\sum_{i=0}^{n-1}(\mu_{\phi}+\delta_{\phi}\cdot i)(\mu_{\Phi}\cdot\delta_{\Phi}^i)$$
-Now, multiply in the two brackets, obtaining: $$\Omega_n=\sum_{i=0}^{n-1}\mu_{\phi}\cdot\mu_{\Phi}\cdot\delta_\Phi ^ i+\delta_\phi\cdot i \cdot\mu_\Phi\cdot\delta_\Phi^i$$
-I can split the four terms into two different sums:$$\Omega_n=\sum_{i=0}^{n-1}\mu_{\phi}\cdot\mu_{\Phi}\cdot\delta_\Phi ^ i+\sum_{i=1}^{n}\delta_\phi\cdot i \cdot\mu_\Phi\cdot\delta_\Phi^i$$
-Being $\mu_\phi,\mu_\Phi,\delta_\Phi,\delta_\phi$ constants, we have: $$\Omega_n=\mu_{\phi}\cdot\mu_{\Phi}\cdot\sum_{i=0}^{n-1}\delta_\Phi ^ i+\delta_\phi\cdot\mu_\Phi\cdot\sum_{i=0}^{n-1} i \cdot\delta_\Phi^i$$
-Let $\Omega_n=\rho_{1,n-1}+\rho_{2,n-1}$
-The first term is a geometric progression, so: $$\rho_{1,n-1}=\mu_{\phi}\cdot\mu_{\Phi}\cdot\frac{1-\delta_\Phi^{n-1}}{1-\delta_\Phi}=\cdot\mu_{\phi}\cdot\mu_{\Phi}\cdot\frac{\delta_\Phi^{n-1}-1}{\delta_\Phi-1}$$
-The second term is again the product of two variable quantities,so here I'm stuck. Any idea of how to move forward?
Thanks.
Ultimately, what we need to evaluate here is a sum of the form $S = \sum_{k=1}^{n-1}kx^k$. That is, I will take $x = \delta_{\Phi}$ and use $k$ as my index of summation; the term added for $i=0$ is $0$, so we can leave it out.
Consider the function $f(x) = \sum_{k=1}^{n-1}kx^{k-1}$ (which is $S/x$). We note that $$ \int_{0}^t f(x)\,dx = \int_0^t \left(\sum_{k=1}^{n-1}kx^{k-1}\right)\,dx = \sum_{k=1}^{n-1} \int_0^t kx^{k-1} \,dx = \sum_{k=1}^{n-1}t^k. $$ Using the formula for the sum of a geometric progression, we have $$ \int_{0}^t f(x)\,dx = t\sum_{k=1}^{n-1}t^{k-1} = t \frac{t^{n}-1}{t-1} = \frac{t^{n+1} - t}{t-1}. $$ Taking the derivative of both sides, we have $$ f(t) = \frac{d}{dt}\left[ \frac{t^k - t}{t-1} \right] = \frac{nt^{n+1} - (n+1)t^n + 1}{(t-1)^2}. $$ So, we finally have $$ \sum_{k=1}^{n-1}kx^k = xf(x) = x \cdot \frac{nx^{n+1} - (n+1)x^n + 1}{(x-1)^2}. $$