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For instance, I know that the sum of the first $101$ natural numbers can be expressed in the following easy computation:

$\sum_{i=1}^{101}i = \frac{101*102}{2}$

One of the questions is: and what about this sum?

$\sum_{i=1}^{101}i + \sum_{i=1}^{100}i + ... + \sum_{i=1}^{1}i = \sum_{i_1=1}^{101}\sum_{i_2=1}^{i_1}i_2$

And specially, what about the $nth$ case, i.e.;

$\sum_{i_1=1}^{101}\sum_{i_2=1}^{i_1}\cdots\sum_{i_n=1}^{i_{n-1}}i_n$

Thanks in advance!

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  • $\begingroup$ how do you plan to apply this to quantitative finance? $\endgroup$ – develarist Nov 29 '19 at 15:10
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    $\begingroup$ I am just extracting means and standard deviations of different portfolios. Suppose we have two assets: $X_1$ and $X_2$ and two ponderations $\alpha_1$ and $\alpha_2$ that must add up to one and each must be between 0 and 1. Then our portfolio is of coarse $\alpha_1 \cdot X_1 + \alpha_2 \cdot X_2$. If we allow the $\alpha's$ to increment by $.01$ then we have 101 possible different portfolios. WIth the same assumptions, but with three assets, we have $\sum_{i=1}^{101}i$ different portfolios. For the $n=4$ case we have $\sum_{i=1}^{101}\sum_{j=1}^{i}j$ possible cases, and so on. $\endgroup$ – Mathias Barreto Nov 29 '19 at 16:55
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As you mentioned, we have

$$\sum_{l=0}^{k}{p}=\frac{k(k+1)}{2}$$

You want to know

$$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\sum_{k=0}^{n}{\frac{k(k+1)}{2}}=\frac{1}{2}\sum_{k=0}^{n}{k^2}+\frac{1}{2}\sum_{k=0}^{n}{k}$$

you know that

$$\sum_{k=0}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}$$

Therefore $$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}=\frac{n(n+1)(n+2)}{6}$$

EDIT :

We can generalize it : For $m$ iterations,summing up to $k$, we have

$$Sum(m,k)=\frac{k(k+1)...(k+m)}{(m+1)!}$$

In other words,

$$Sum(m,k)={m+k \choose k-1}$$

$$Sum(m+1,n)=\sum_{k=0}^{n}{{m+k \choose k-1}}$$

Also,

$${m+1 +n \choose n-1}={m+n\choose n-2}+{m+n\choose n-1}$$ $${m+n\choose n-2}={m+n-1\choose n-3}+{m+n-1\choose n-2}$$ We keep doing this, until we get $$Sum(m+1,n)={m+1 +n \choose n-1}$$

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  • $\begingroup$ I conjecture that the next one is $\frac{n(n+1)(n+2)(n+3)}{4!}$ $\endgroup$ – Alex C Nov 29 '19 at 16:48
  • $\begingroup$ Please prove it :D $\endgroup$ – Mathias Barreto Nov 29 '19 at 17:09
  • $\begingroup$ Alex C is correct $\endgroup$ – Canardini Nov 29 '19 at 18:37
  • $\begingroup$ @Canardini It worked for the first two terations ($m=2$), but just wondering, how did you come up with the answer? Did you prove it by induction? $\endgroup$ – Mathias Barreto Dec 3 '19 at 12:39
  • $\begingroup$ With m=1 et m=2, we can clearly see a pattern. Yes it is proved by induction. If you believe that the post answers your question, please accept the answer $\endgroup$ – Canardini Dec 3 '19 at 13:43

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