Coefficient of expansion and limits

Let A n { A }_{ n } be the coefficient of x n + 3 {x}^{-n+3} in the expansion of ( x + 1 x ) n + 1 \left(x+ \frac {1}{x} \right)^{n+1} . Determine the value of n = 1 1 A n , \displaystyle{\sum _{ n=1 }^{ \infty }}{ \frac { 1 }{ { A }_{ n } } } , where n n is a positive integer.


The answer is 2.

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1 solution

Avi Aryan
Jun 3, 2014

When x=1 , coefficient of x^2 = 2c2 = 1
When x=2 , coefficient of x^1 = 3c2 = 3
When x=3 , coefficient of x^0 = 4c2 = 6
When x=4 , coefficient of x^-1 = 5c3 = 10
When x=5 , coefficient of x^-2 = 6c4 = 15
When x=6 , coefficient of x^-3 = 7c5 = 21


Here you can see that for x = n u m x = num , the coefficient is sum of first n u m num natural numbers.
So when x = 7 , the coefficient = 8 c 6 = 8 7 / 2 = 28 = 8c6 = 8*7/2 = 28

General term for the reciprocal 1 A n = 1 n ( n + 1 ) / 2 \frac{1} A_{n} = \frac{1} {n(n+1)/2}
General term 1 A n = 2 ( 1 n 1 n + 1 ) \frac{1} A_{n} = 2( \frac{1}{n} - \frac{1}{n+1} )

The sum = 2 ( ( 1 1 1 2 ) + ( 1 2 1 3 ) + ( 1 3 1 4 ) + 1 ) 2( ( \frac{1}{1} - \frac{1}{2} ) + ( \frac{1}{2} - \frac{1}{3} ) + ( \frac{1}{3} - \frac{1}{4} ) + \ldots - \frac{1}{ \infty } ) = 2 1 2*1

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