Binomially Amazing!

Calculus Level 5

$(1+x)^{1729} = C_0 + C_1 x +C_2 x^2 + \ldots + C_{1729} x^{1729}$

The equation above is an algebraic identity for constants $C_0,C_1,\ldots,C_{1729}$ . Given that

$\frac{C_0}1 - \frac{C_1}5 + \frac{C_2}9 - \ldots - \frac{C_{1729}}{6917}$

is equal to $\dfrac{a! \ 4^b}{c!!!!}$ , where $a,b$ and $c$ are integers, find the value of $a+b+c$ .

Clarification :

$n!!!!$ denotes the quadruple factorial, eg $15!!!! = 15\times11\times7 \times3$ .
$1,5,9,\ldots,6917$ follows an arithmetic progression.

The answer is 10375.

2 solutions

Aditya Kumar
Nov 11, 2015

$\displaystyle { \left( 1-{ x }^{ 4 } \right) }^{ 1729 }={ C }_{ 0 }-{ C }_{ 1 }x^{ 4 }+{ C }_{ 2 }x^{ 8 }-{ C }_{ 3 }x^{ 12 }+...-{ C }_{ 1729 }x^{ 6916 }\\ \int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 4 } \right) }^{ 1729 }dx } =\int _{ 0 }^{ 1 }{ \left( { C }_{ 0 }-{ C }_{ 1 }x^{ 4 }+{ C }_{ 2 }x^{ 8 }-{ C }_{ 3 }x^{ 12 }+...-{ C }_{ 1729 }x^{ 6916 } \right) } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { { C }_{ 0 } }{ 1 } -\frac { { C }_{ 1 } }{ 5 } +\frac { { C }_{ 2 } }{ 9 } -...-\frac { { C }_{ 1729 } }{ 6917 } \quad \quad \quad \quad \quad \quad \quad \quad \quad .......(1)\\ I=\int _{ 0 }^{ 1 }{ { \left( 1-{ x }^{ 4 } \right) }^{ 1729 }dx } \\ Let\quad { x }^{ 4 }=t\\ \therefore \quad I=\frac { 1 }{ 4 } \int _{ 0 }^{ 1 }{ { { t }^{ \frac { -3 }{ 4 } }\left( 1-t \right) }^{ 1729 }dx } \\ \quad \quad \quad =\frac { 1 }{ 4 } \frac { \Gamma \left( 1730 \right) \Gamma \left( \frac { 1 }{ 4 } \right) }{ \Gamma \left( \frac { 6921}{ 4 } \right) } \quad \quad \quad \quad \quad (Using\quad Beta\quad Function)\\ \quad \quad \quad =\quad \frac { 1 }{ 4 } \frac { \Gamma \left( 1730 \right) \Gamma \left( \frac { 1 }{ 4 } \right) }{ \left( \frac { 6917 }{ 4 } \right) \left( \frac { 6913 }{ 4 } \right) \left( \frac { 6909 }{ 4 } \right) ...\left( \frac { 5 }{ 4 } \right) \left( \frac { 1 }{ 4 } \right) \Gamma \left( \frac { 1 }{ 4 } \right) } \\ \quad \quad \quad =\frac { 1729!4^{ 1729 } }{ 6917!!!! } \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ......(2)\\ On\quad equating\quad equation\quad 1\& 2,\\ \frac { { C }_{ 0 } }{ 1 } -\frac { { C }_{ 1 } }{ 5 } +\frac { { C }_{ 2 } }{ 9 } -...-\frac { { C }_{ 1729 } }{ 6917 } =\frac { 1729!4^{ 1729 } }{ 6917!!!! } \\$

I too did this exactly in the same way.

Surya Prakash - 5 years, 7 months ago

Nice I just forgot to do this way..one of the beautiful solutions to this elegant question!...my question is if it is possible to do this one by generalising the rth term and then apply the summation operator??? @Pi Han Goh and @Aditya Kumar !! Help me!

Righved K - 5 years, 7 months ago

See Jason's solution.

Pi Han Goh - 5 years, 7 months ago

Thanku @Pi Han Goh for ur reply but stillthat solution is not catering my problem....I thought whether we could do some algebraic manipulations in the generalised term of the above question and if some sort of series come out without any use of calculus... I hope u understand my problem cheers!

Righved K - 5 years, 7 months ago

@Righved K – I don't think that's possible and I doubt that it's possible to solve it without calculus.

Pi Han Goh - 5 years, 4 months ago

@Pi Han Goh is it fine to post it in algebra?

Aditya Kumar - 5 years, 7 months ago

Nice question did not saw we could have integrated the question.

Department 8 - 5 years, 7 months ago

Thanks! The denominators give a big hint of the pattern formed. Eg. Integral of x^4 gives a denominator of 5.

Aditya Kumar - 5 years, 7 months ago

@Aditya Kumar – LUV THIS QUESTION! THANKS! It should be in calculus.

Pi Han Goh - 5 years, 7 months ago

@Pi Han Goh – All my questions end up in calculus :P

Aditya Kumar - 5 years, 7 months ago

@Aditya Kumar (you're the same Aditya right? Why deactivating?)

Now, as you have used multi-factorial quite well, in probably making this problem. I believe that you can answer Calculus Challenge 10 and prove the general theorem. Hint is given in your problem only.

Kartik Sharma - 5 years, 7 months ago

I hope that this concept is much beyond jee can any one suggest some other method

Gauri shankar Mishra - 5 years, 4 months ago

Read the other solution.

Pi Han Goh - 5 years, 4 months ago

There's another way which uses Ramanujan's Master Theorem.

Aditya Kumar - 5 years, 4 months ago

Thanks dud

Gauri shankar Mishra - 5 years, 4 months ago

@Gauri shankar Mishra – I'm not a dud.

Aditya Kumar - 5 years, 4 months ago

@Aditya Kumar – Sorry Typographical error

Gauri shankar Mishra - 5 years, 4 months ago

Jason Martin
Nov 13, 2015

Note that $(1-x^4)^{1729}=C_0-C_1x^4+C_2x^8- \cdots -C_{1729}x^{6916}$ , so $\int_0^1 (1-x^4)^{1729} dx=\frac{C_0}{1}-\frac{C_1}{5}+\frac{C_2}{9}- \cdots \frac{C_{1729}}{6917}$ , which is the desired sum.

Now define $I(n)= \int_0^1 (1-x^4)^n dx$ . We wish to find a closed form for $I(n)$ by first representing it recursively:

Thus, $I(n)= \int_0^1 (1-x^4)^n dx= \int_0^1 (1-x^4)(1-x^4)^{n-1} dx=\int_0^1 (1-x^4)^{n-1} dx - \int_0^1 x^4(1-x^4)^{n-1} dx=$ $I(n-1) - \int_0^1 x \cdot x^3(1-x^4)^{n-1} dx=I(n-1)+\frac{1}{4n} \int_0^1 (x)(-4nx^3)(1-x^4)^{n-1} dx=$ $I(n-1)+0 -\frac{1}{4n}\int_0^1 (1-x^4)^n dx=I(n-1)-\frac{1}{4n}I(n)$ . Therefore, $I(n)=\frac{4n}{4n+1}I(n-1)$ .

Since $I(0)=1$ , we know the closed form for $I(n)$ is $I(n)=\frac{(4n)(4n-4)(4n-8) \cdots 4}{(4n+1)(4n-3)(4n-7) \cdots 5}=\frac{n!4^n}{(4n+1)!!!!}$ .

For $n=1729$ , we have $a=1729, b=1729, c=6917$ , so the answer is $10375$ .

Nice Solution! Never thought of recurrance :)

Aditya Kumar - 5 years, 6 months ago

Thanks, I based it off of classical techniques for solving integrals because I was less familiar with things like the Gamma function.

Jason Martin - 5 years, 6 months ago

Binomially Amazing!

The answer is 10375.

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