This sum will give you nightmares!!!

In my entry to Troll King Contest, I posed the following problem -

Evaluate

\[\sum_{a=0}^\infty \sum_{b=0}^\infty \sum_{c=0}^a \sum_{d=0}^b \sum_{e=0}^d {n+a \choose a} {b-e \choose d-e} {a \choose c} {b \choose e} (-1)^{b+c} n^e c^{d-e}\]

Let's see who can dare to attempt it.

I'll post my solution, once the troll contest ends.

Also notice that, this problem inspite of an entry in Toll King Contest serves another purpose of explaining a little bit about multiple summations to my friends who requested it.

[Hint for solving that troll problem : After reading the complete problem, ask yourself that do I really need to troll with the above summation]

#Series #Summation #Troll #Kishlaya #MultipleSummation

Note by Kishlaya Jaiswal
6 years, 2 months ago

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1 vote

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Comments

Answer as per me is 11

Since the sum is large and solution long, so i shall write only the usable parts of the sum while writing it, please bear with me

Let us first consider the summation of 'e' terms

it can be shown with some manipulation that (as i have told, the other terms have been ommitted for clarity)

∑(b−ed−e)(be)(nce)=∑(bd)(de)(nce)=(bd)(1+nc)d\sum { (\begin{matrix} b-e \\ d-e \end{matrix})(\begin{matrix} b \\ e \end{matrix})({ \frac { n }{ c } }^{ e }) } \quad =\quad \quad \sum { (\begin{matrix} b \\ d \end{matrix})(\begin{matrix} d \\ e \end{matrix})({ \frac { n }{ c } }^{ e }) } =(\begin{matrix} b \\ d \end{matrix})(1+\frac { n }{ c } )^{ d }

where the summation is over 'e'

now , let us consider the sum over 'd'

we have the 'd' having terms as (now)

∑(bd)(1+nc)dcd=∑(bd)(n+c)d=(1+n+c)b\sum { (\begin{matrix} b \\ d \end{matrix})(1+\frac { n }{ c } )^{ d }{ c }^{ d } } \quad =\quad \sum { (\begin{matrix} b \\ d \end{matrix})(n+c)^{ d } } =(1+n+c)^{ b }

now, since the summation over 'b' is not related to that over 'c' in any way, so we perform the sum over 'b' first to write

consider all 'b' terms

∑(−1)b(1+n+c)b=12+n+c(fromtaylorseriesof11+xatx=1+n+c)\sum { (-1)^{ b } } (1+n+c)^{ b }\quad =\quad \frac { 1 }{ 2+n+c } \quad (from\quad taylor\quad series\quad of\quad \frac { 1 }{ 1+x } \quad at\quad x=\quad 1+n+c\quad )

where summation was over 'b'

now , we have to sum over 'c' to 'a' first and then 'a' to ∞\infty

so consider all 'c' terms presently there

∑(−1)c2+n+c(ac)\sum { \frac { (-1)^{ c } }{ 2+n+c } } (\begin{matrix} a \\ c \end{matrix})\quad

to solve it , consider

(1−x)a=1−(aC1)x+(aC2)x2........(1-x)^{ a }\quad =\quad 1-\quad (aC1)x+(aC2){ x }^{ 2 }........

multiply by xn+1{ x }^{ n+1 }\quad and integrate from 0 to 1 to get the desired summation (check yourself)

as

∫01(1−x)axn+1dx=a!(n+1)!(a+n+2)!\int _{ 0 }^{ 1 }{ (1-x)^{ a } } { x }^{ n+1 }dx\quad =\quad \frac { a!(n+1)! }{ (a+n+2)! }

where the integral has been evaluated using beta function,

finally we have the summation over 'a' as

(a+nn)a!(n+1)!(a+n+2)!=(a+n)!a!(n+1)!a!n!(a+n+2)!=n+1(a+n+1)(a+n+2)(\begin{matrix} a+n \\ n \end{matrix})\frac { a!(n+1)! }{ (a+n+2)! } =\frac { (a+n)!a!(n+1)! }{ a!n!(a+n+2)! } =\frac { n+1 }{ (a+n+1)(a+n+2) }

hence

∑n+1(a+n+1)(a+n+2)=∑(1a+n+1−1a+n+2)(n+1)=1(telescopicseries)\sum { \frac { n+1 }{ (a+n+1)(a+n+2) } } \quad =\quad \sum { (\frac { 1 }{ a+n+1 } -\frac { 1 }{ a+n+2 } )(n+1) } \quad =\quad 1\quad (telescopic\quad series)\\

where summation is over 'a'

hence answer is 1\displaystyle 1

Kishlaya Jaiswal Am i right?

Mvs Saketh - 6 years, 2 months ago

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Yeah right!

I have got even a simpler solution. I'll share it after the Troll Contest Ends (or maybe earlier, only if I see that I have got too less votes = upvotes + downvotes\sqrt{\text{downvotes}})

Kishlaya Jaiswal - 6 years, 2 months ago

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alright, beautifully moulded question btw +1

Mvs Saketh - 6 years, 2 months ago
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