Is it possible to put series in other series?

Question

I've been working on a project for quite a long time but I found myself stuck at a step where I have to multiply elements of a series by elements of another series, which is dependent on the former one. Eventually, it should look like this: $$\sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\sum_{c=1}^i(2c-1)$$ where every result of the first series multiplies the various results of the second one, whose number increase with the value of i.

I would like to know if it is possible, as I haven't ever seen anything like this (I am little more than an amateur at Maths), and if it is not, I'd really appreciate some suggestions on alternative ways to do it.

Thanks in advance!

Answer 1

Hint. One may recall that $$ \sum_{c=1}^i(2c-1)=i^2 $$ and that $$ \sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6, \quad \quad \sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4. $$

Thus your initial sum rewrites $$ \sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\sum_{c=1}^i(2c-1)=\sum_{i=1}^{\lfloor\frac{(n_f-11)}6\rfloor}(2i+3)\cdot i^2=2\sum_{i=1}^{n}i^3+3\sum_{i=1}^{n}i^2 $$ with $\displaystyle n=\lfloor\frac{(n_f-11)}6\rfloor$.

Answer 2

Notice that the inner sum is equal to

$$2 \sum \limits _{c=1} ^i c - \sum \limits _{c=1} ^i 1 = 2 \frac {i(i+1)} 2 - i = i^2 .$$

Your sum, then, becomes $$\sum \limits _{i=1} ^N (2i+3) i^2 = 2 \sum \limits _{i=1} ^N i^3 + 3 \sum \limits _{i=1} ^N i^2 = 2 \frac {N^2 (N+1)^2} 4 + 3 \frac {N(N+1)(2N+1)} 6 = \frac {N^4 + 4N^3 + 4N^2 + N} 2 ,$$

where $N = \lfloor \dfrac {n_f-11} 6 \rfloor$.

The sums of powers that I have used are well known, you can find a brief list on Wikipedia.