An algebra problem by Rocco Dalto

Algebra Level 4

Let M M be any positive real number.

If J = 1 1000 1 ( X J ) ( X J 1 ) = 1 M \displaystyle \sum_{J=1}^{1000} \dfrac1{(X-J)(X-J-1)} = \dfrac1M , find the sum of roots of X X .


The answer is 1002.

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1 solution

Rocco Dalto
Sep 1, 2016

F o r For a n y any p o s i t i v e positive i n t e g e r integer J J u s i n g using p a r t i a l partial f r a c t i o n s fractions w e we h a v e : have: 1 ( X J ) ( X J 1 ) = \frac{1}{(X - J) * (X - J -1)} = A X J + B X J 1 \frac{A}{X - J} + \frac{B}{X - J - 1} \implies
A + B = 0 A + B = 0 ( J + 1 ) A + J B = 1 (J + 1) * A + J * B = -1 \implies A = 1 A = -1 a n d and B = 1 B = 1 \implies 1 ( X J ) ( X J 1 ) = \frac{1}{(X - J) * (X - J -1)} = 1 X J 1 1 X J \frac{1}{X - J - 1} - \frac{1}{X - J} \therefore f o r for a n y any p o s i t i v e positive i n t e g e r integer N N w e we h a v e : have: j = 1 N 1 ( X J ) ( X J 1 ) \sum\limits_{j = 1}^N \frac{1}{(X - J) * (X - J -1)} = = 1 X N 1 1 X 1 \frac{1}{X - N - 1} - \frac{1}{X - 1} = 1 M , \frac{1}{M}, w h e r e where M > 0. M > 0.

F o r For M > 0 M > 0 w e we h a v e have t h e the r e a l real s o l u t i o n s : solutions:

X 1 , X 2 = ( N + 2 ) ± N ( N + 4 M ) 2 X_1,X_2 = \frac{(N + 2) \pm \sqrt{N * (N + 4 * M)}}{2} a n d and X 1 + X 2 = N + 2 X_1 + X_2 = N + 2 \therefore N = 1000 N = 1000 \implies X 1 + X 2 = 1002. X_1 + X_2 = 1002.

Shouldn't A+B be equals to 1?

Pablo Esteban Salinas Solis - 4 years, 9 months ago

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A + B = 0 , A + B = 0, s i n c e since w h e r e where e q u a t i n g equating c o e f f i c i e n t s . coefficients.

1 = A ( X J 1 ) + B ( X J ) = ( A + B ) X ( ( J + 1 ) A + B J ) . 1 = A * (X - J - 1) + B * (X - J) = (A + B) * X - ((J + 1) * A + B * J).

E q u a t i n g Equating c o e f f i c i e n t s coefficients w e we o b t a i n : obtain:

A + B = 0 A + B = 0

( J + 1 ) A + J B = 1 (J + 1) * A + J * B = -1

Rocco Dalto - 4 years, 9 months ago

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