Recurring Recurrence

Algebra Level 5

Find $ax^5 + by^5$ if the real numbers $a, b, x$ , and $y$ satisfy the equations $ax + by = 3$ $ax^2 + by^2 = 7$ $ax^3 + by^3 = 16$ $ax^4 + by^4 = 42$

Try my set .

The answer is 20.

1 solution

Parth Lohomi
Apr 9, 2015

A recurrence of the form $T_n=AT_{n-1}+BT_{n-2}$ will have the closed form $T_n=ax^n+by^n$ , where $x,y$ are the values of the starting term that make the sequence geometric, and $a,b$ are the appropriately chosen constants such that those special starting terms linearly combine to form the actual starting terms.

Suppose we have such a recurrence with $T_1=3$ and $T_2=7$ . Then $T_3=ax^3+by^3=16=7A+3B$ , and $T_4=ax^4+by^4=42=16A+7B$ .

Solving these simultaneous equations for $A$ and $B$ , we see that $A=-14$ and $B=38$ . So, $ax^5+by^5=T_5=-14(42)+38(16)= \boxed{20}$

Maybe Parth said all this, and I missed some of it. Here is how I solved it: Starting with: $ax^3 + by^3 = 16$ , substitute for $ax^2$ and $by^2$ from previous equation.

$x (7- by^2) + y (7- ax^2) = 16$

$7(x+y) - (ax + by) xy = 16$

$7(x+y) -3xy = 16$

Performing similar math on fourth equation yields: $x(16-by^3) + y(16-ax^3) = 42$

$16(x+y) - (ax^2 + by^2)xy = 42$

$16(x+y) - 7xy = 42$

This is now two linear equations in two unknowns. Let $S = x + y$ , let $T = xy$ . Now you have the two equations that Parth solved.

$7S - 3T = 16$

$16S - 7T = 42$

Solve for $S$ and $T$ as above, and use same pattern of recurrence to calculate final answer.

Bradley Slavik - 6 years, 1 month ago

Recurring Recurrence

Try my set .

The answer is 20.

1 solution

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