f ( n ) = 1 ( x + n ) ( x + n + 1 ) f(n)=\frac{1}{(x+n)(x+n+1)}

Algebra Level 2

f ( n ) = 1 ( x + n ) ( x + n + 1 ) f(n)=\frac{1}{(x+n)(x+n+1)} for a positive integer n n . If the following always holds regardless of the value of x x , what is the sum of the constants a + b + c a+b+c : f ( 12 ) + f ( 13 ) + f ( 14 ) + ⋯ + f ( 19 ) = c ( x + a ) ( x + b ) ? f(12)+f(13)+f(14)+\cdots+f(19)=\frac{c}{(x+a)(x+b)}?

40 37 34 31

Ajay Dutta by Ajay Dutta , 24, India

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2 solutions

f ( n ) = 1 ( x + n ) ( x + n + 1 ) f(n) = \dfrac{1}{(x+n)(x+n+1)}

f ( n ) = ( x + n + 1 ) − ( x + n ) ( x + n ) ( x + n + 1 ) f(n) = \dfrac{(x+n+1) - (x+n)}{(x+n)(x+n+1)}

f ( n ) = 1 x + n − 1 x + n + 1 f(n) = \dfrac{1}{x+n} - \dfrac{1}{x+n+1}

⇒ f ( 12 ) + f ( 13 ) + ⋯ + f ( 18 ) + f ( 19 ) = 1 x + 12 − 1 x + 13 + 1 x + 13 − 1 x + 14 + … + 1 x + 18 − 1 x + 19 + 1 x + 19 − 1 x + 20 \Rightarrow f(12) + f(13) + \dots + f(18) + f(19) = \frac{1}{x + 12} - \frac{1}{x+13} + \frac{1}{x + 13} - \frac{1}{x+14} + \ldots + \frac{1}{x + 18} - \frac{1}{x+19} + \frac{1}{x + 19} - \frac{1}{x+20}

= 1 x + 12 − 1 x + 20 = \dfrac{1}{x + 12} - \dfrac{1}{x+20}

= ( x + 20 ) − ( x + 12 ) ( x + 12 ) ( x + 20 ) = \dfrac{(x+20) - (x+12)}{(x+12)(x+20)}

= 8 ( x + 12 ) ( x + 20 ) = \dfrac{8}{(x+12)(x+20)}

Therefore, a + b + c = 8 + 12 + 20 = 40 a + b + c = 8 + 12 + 20 = \boxed{40}

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Ravi Bendi
Mar 21, 2014

same as "Siddhartha Srivastava" solution

0 Helpful 0 Interesting 0 Brilliant 0 Confused

I am not expert in math. But try to understand problems. Is there any rules applied?

Samar Cxma - 7 years, 2 months ago

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