1990 AIME Problema 15

Question

Halla

ax^5+by^5

si los números reales

a,

b,

x,

y

y

satisfacen las ecuaciones

\begin{aligned}ax+by&=3,\\ax^2+by^2&=7,\\ax^3+by^3&=16,\\ax^4+by^4&=42.\end{aligned}

Find

ax^5+by^5

if the real numbers

a,

b,

x,

and

y

satisfy the equations

\begin{aligned}ax+by&=3,\\ax^2+by^2&=7,\\ax^3+by^3&=16,\\ax^4+by^4&=42.\end{aligned}

Respuesta

Respuesta: 20

Conceptos:recursión sistema de ecuaciones simetría (álgebra)

Nivel de dificultad: 2270

Pista pequeña:

Sea

S_k=ax^k+by^k

y deriva una recurrencia usando

x+y

y

xy

Let

S_k=ax^k+by^k

and derive a recurrence using

x+y

and

xy

Pista grande:

Usa

S_3

y

S_4

para hallar los dos coeficientes de la recurrencia antes de calcular

S_5

Use

S_3

and

S_4

to solve for the two recurrence coefficients before computing

S_5

Solución:

Sean

S_k=ax^k+by^k,

p=x+y,

y

q=xy.

Como

x

y

y

satisfacen

t^2=pt-q,

S_{k+2}=pS_{k+1}-qS_k.

Al usar

S_1=3,

S_2=7,

S_3=16,

y

S_4=42

se obtiene

\begin{aligned}7p-3q&=16,\\16p-7q&=42.\end{aligned}

Al resolver,

p=-14

y

q=-38.

Por lo tanto,

\begin{aligned}S_5&=pS_4-qS_3\\&=-14(42)+38(16)\\&=20.\end{aligned}

Let

S_k=ax^k+by^k,

p=x+y,

and

q=xy.

Since

x

and

y

each satisfy

t^2=pt-q,

S_{k+2}=pS_{k+1}-qS_k.

Using

S_1=3,

S_2=7,

S_3=16,

and

S_4=42

gives

\begin{aligned}7p-3q&=16,\\16p-7q&=42.\end{aligned}

Solving yields

p=-14

and

q=-38.

Therefore

\begin{aligned}S_5&=pS_4-qS_3\\&=-14(42)+38(16)\\&=20.\end{aligned}