Are you game for gamma?

Probability Level 4

I f ( 1 + x ) n = C 0 + C 1 x + C 2 x 2 + C 3 x 3 + . . . + C n x n , t h e n t h e v a l u e o f C 0 n − C 1 n + 1 + C 2 n + 2 − . . . . . + ( − 1 ) n C n 2 n = Γ ( x ) Γ ( y ) Γ ( z ) . F i n d ( z − x − y ) 2 n 2 If\quad { (1+x) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }{ x }+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...+{ C }_{ n }{ x }^{ n },\quad then\quad the\quad value\quad of\quad \\ \frac { { C }_{ 0 } }{ n } -\frac { { C }_{ 1 } }{ n+1 } +\frac { { C }_{ 2 } }{ n+2 } -.....+{ (-1) }^{ n }\frac { { C }_{ n } }{ 2n } =\frac { \Gamma (x)\Gamma (y) }{ \Gamma (z) } .\\ Find\quad (z-x-y)2{n}^{2}


The answer is 0.

Vishnu C by vishnu c , 22, India

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1 solution

Vishnu C
Mar 20, 2015

Just for the record, I find the fact that for functions like the gamma function, Γ ( n ) = ( n − 1 ) ! \Gamma(n)=(n-1)! quite annoying. This problem is quite straight forward - multiply ( 1 + x ) n b y x n {(1+x)}^{n} \quad by \quad {x}^{n} ; then the integration should be pretty easy. You should get ( n − 1 ) ! ( n ) ! ( 2 n ) ! = B ( n , n + 1 ) . ∴ z = 2 n + 1 ; x = n o r n + 1 a n d y = n + 1 o r n r e s p e c t i v e l y . ⇒ z − x − y = 0 \frac { (n-1)!(n)! }{ (2n)! } =B(n,n+1).\\ \therefore z=2n+1;\quad x=n\quad or\quad n+1\quad and\quad y=n+1\quad or\quad n\quad respectively.\\ \Rightarrow \quad z-x-y=\boxed { 0 } .

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Yes B ( m , n ) T ( m + n ) = T ( m ) T ( n ) B(m,n)T(m+n)=T(m)T(n) .Holds if one guesses to be a direct form of the beta function(that's quite apparent though)its easy that z = x + y z=x+y

Spandan Senapati - 4 years, 1 month ago

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