Autor Tema: Binomial sum

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24 Abril, 2019, 07:54 pm
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jacks

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Sum of series  \( \displaystyle \sum^{n}_{k=1}\sum^{n}_{j=1}\sum^{n}_{i=1}\bigg[\binom{n}{i}\bigg(\binom{n}{j}-\binom{n-i}{j}\bigg)\bigg(\binom{n}{k}-\binom{n-j}{k}\bigg)\bigg] \)

25 Abril, 2019, 10:45 am
Respuesta #1

Luis Fuentes

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Hi

Sum of series  \( \displaystyle \sum^{n}_{k=1}\sum^{n}_{j=1}\sum^{n}_{i=1}\bigg[\binom{n}{i}\bigg(\binom{n}{j}-\binom{n-i}{j}\bigg)\bigg(\binom{n}{k}-\binom{n-j}{k}\bigg)\bigg] \)

Note that:

\( \displaystyle\sum_{i=0}^n{}\displaystyle\binom{n}{i}a^ib^{n-i}=(a+b)^n \)

\( \displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}a^ib^{n-i}=(a+b)^n-b^n \)

\( \displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}a^i=(a+1)^n-1 \)

\( \displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}b^{n-i}=(b+1)^n-b^n \)

From this:

(1) \( \displaystyle\sum_{i=1}^n{}\displaystyle\sum_{j=1}^n{}\displaystyle\sum_{k=1}^n{}\displaystyle\binom{n}{i}\displaystyle\binom{n}{j}\displaystyle\binom{n}{k}=\left(\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\right)\left(\displaystyle\sum_{j=1}^n{}\displaystyle\binom{n}{j}\right)\left(\displaystyle\sum_{k=1}^n{}\displaystyle\binom{n}{k}\right)=(2^n-1)^3 \)

(2) \( \displaystyle\sum_{i=1}^n{}\displaystyle\sum_{j=1}^n{}\displaystyle\sum_{k=1}^n{}\displaystyle\binom{n}{i}\displaystyle\binom{n}{j}\displaystyle\binom{n-j}{k}=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^n{}\displaystyle\binom{n}{j}\displaystyle\sum_{k=1}^{n-j}\displaystyle\binom{n-j}{k}=\\=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^n{}\displaystyle\binom{n}{j}(2^{n-j}-1)=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}(3^n-2^n-2^n+1)=(2^n-1)(3^n-2^{n+1}+1) \)

(3) \( \displaystyle\sum_{i=1}^n{}\displaystyle\sum_{j=1}^n{}\displaystyle\sum_{k=1}^n{}\displaystyle\binom{n}{k}\displaystyle\binom{n}{i}\displaystyle\binom{n-i}{j}=\ldots=(2^n-1)(3^n-2^{n+1}+1) \)

(4) \( \displaystyle\sum_{i=1}^n{}\displaystyle\sum_{j=1}^n{}\displaystyle\sum_{k=1}^n{}\displaystyle\binom{n}{i}\displaystyle\binom{n-i}{j}\displaystyle\binom{n-j}{k}=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^n\displaystyle\binom{n-i}{j}\displaystyle\sum_{k=1}^{n-j}\displaystyle\binom{n-j}{k}=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^n\displaystyle\binom{n-i}{j}(2^{n-j}-1)=\\
=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^{n-i}\displaystyle\binom{n-i}{j}(2^{n-j}-1)=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}\displaystyle\sum_{j=1}^{n-i}\displaystyle\binom{n-i}{j}(2^i\cdot 2^{n-i-j}-1)=\\
=\displaystyle\sum_{i=1}^n{}\displaystyle\binom{n}{i}(2^i(3^{n-i}-2^{n-i})-(2^{n-i}-1))=\ldots=
5^n-3^n-2^n(2^n-1)-(3^n-2^n)+(2^n-1) \)

Now your sum is:

\( (1)-(2)-(3)+(4)=\ldots=8^n+5^n-3^n2^{n+1} \)

Best regards.

27 Abril, 2019, 04:37 pm
Respuesta #2

jacks

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Thanks Admin Got it.