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!

  • $\begingroup$ Is the last $-1$ part of the summand? $\endgroup$ – GFauxPas Jul 17 '16 at 18:00
  • $\begingroup$ Yes, it is! The result of the first series (should) multiply each result of the second one $\sum_{c=1}^i2c-1$ $\endgroup$ – Estagon Jul 17 '16 at 18:37
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    $\begingroup$ Then it should be in paretheses $\endgroup$ – GFauxPas Jul 17 '16 at 18:38
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    $\begingroup$ Edited. Thanks for pointing that out! $\endgroup$ – Estagon Jul 17 '16 at 18:39

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.


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 $\displaystyle n=\lfloor\frac{(n_f-11)}6\rfloor$.

  • $\begingroup$ @Estagon Did my above hint help you? Thanks. $\endgroup$ – Olivier Oloa Jul 17 '16 at 18:51

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