Behold! Alpha and Beta are on their way!

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Let α n \alpha_{n} and β n \beta_{n} be the roots of

x 2 + ( 2 n + 1 ) x + n 2 x^2+(2n+1)x+n^2

The sum of

1 ( α 3 + 1 ) ( β 3 + 1 ) + 1 ( α 4 + 1 ) ( β 4 + 1 ) + … + 1 ( α 20 + 1 ) ( β 20 + 1 ) \frac{1}{(\alpha_{3}+1)(\beta_{3}+1)}+\frac{1}{(\alpha_{4}+1)(\beta_{4}+1)}+\ldots+\frac{1}{(\alpha_{20}+1)(\beta_{20}+1)}

can be expressed as a b \frac{a}{b} , where a a and b b are coprime positive integers.

What is the value of b − a b-a ?


The answer is 229.

Jason Sebastian by Jason Sebastian , 21, Indonesia

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1 solution

Notice that α n β n = n 2 \alpha_{n} \beta_{n} = n^{2} and α n + β n = − ( 2 n + 1 ) \alpha_{n} + \beta_{n} = -(2n+1) from Vieta's formula.

Consider the expression ( α n + 1 ) ( β n + 1 ) (\alpha_{n}+1)(\beta_{n}+1) .

= α n β n + α n + β n + 1 = \alpha_{n} \beta_{n} + \alpha_{n} + \beta_{n} + 1

= n 2 − 2 n = n^{2} - 2n

= n ( n − 2 ) = n(n-2) ...

Therefore, 1 ( α n + 1 ) ( β n + 1 ) \frac{1}{(\alpha_{n}+1)(\beta_{n}+1)}

= 1 n ( n − 2 ) = \frac{1}{n(n-2)}

= 1 2 ( 1 n − 1 n − 2 ) = \frac{1}{2} (\frac{1}{n} - \frac{1}{n-2})

The summation from n = 3 n = 3 to 20 20 is...

= 1 2 ( 1 + 1 2 − 1 19 − 1 20 ) = \frac{1}{2} (1 + \frac{1}{2} - \frac{1}{19} - \frac{1}{20})

= 531 760 = \frac{531}{760}

The answer is 760 − 531 = 229 760 - 531 = \boxed{229} ~~~

Confession: I kept adding 2 numbers together and not realized until I looked at the question... T__T

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