Let $\alpha$ and $\beta$ be the roots of $x^2-x-1=0$. If $P_k=(\alpha)^k+(\beta)^k$, $k\ge 1$, then prove that-

Question

a) $P_5=11$$

b) $P_1+P_2+P_3+P_4+P_5 =26$

For the first part $$\alpha^5+\beta ^5$$ $$=(\alpha^3+\beta ^3)^2-2(\alpha \beta )^3$$

I found the value of $\alpha^3+\beta^3=4$

So $$16-2(-1)=18$$ which doesn’t match.

In the second part depends on the value obtained from part 1, so I need to get that cleared up.

I checked the computation many times, but it might end up being just that. Also, is there a more efficient way to do this?

Answer 1

if $$x^2-x-1=0~~~(1)$$ has roots as $a,b$ then $P_k=a^k+b^k,P_0=2,P_1=a+b=1$ and $$a^2-a-1=0~~~(2),~~ b^2-b-1=0~~~(3)$$ Multiply Eq.(2) once by $a^k$ and (3) by $b^k$. Adding these two Eqs. you get $$(a^{k+2}+b^{k+2})-(a^{k+1}+b^{k+1})-(a^k+b^k)=0 \implies P_{k+2}= P_{k+1}+P_k~~~(4)$$ So $P_5=P_4+P_3, P_4=P_3+P_2, P_3=P_2+P_1, P_2=P_1+P_0=a+b+2=3$$ $$ \implies P_3=3+1=4, \implies P_4=4+3=7, P_5=7+4=11$ So $P_1+P_2+P_3+P_4+P_5=1+3+4+7+11=26$

Answer 2

Hint:

Use $$(a^3+b^3)(a^2+b^2)=a^5+b^5+a^2b^2(a+b)$$ as we know $a+b, ab$

Answer 3

You can make life easier by realizing that

$$P_k = \alpha^k+\beta^k$$

is the solution of the linear recurrence

$$a_{k+2}=a_{k+1}+a_k \text{ with } a_1 = \alpha + \beta = 1 $$ $$\text{ and } a_2 = \alpha^2 + \beta^2 = (\alpha + \beta)^2-2\alpha\beta = 1+2=3$$

Hence,

$$a_3= 4, a_4 = 7, a_5 = 11$$

Answer 4

First, prove the identity $$P_{k+1} = P_1 P_k - \alpha\beta P_{k-1}.$$ Then note that if $\alpha, \beta$ are roots of $x^2 - x - 1$, then they admit the factorization $$x^2 - x - 1 = (x-\alpha)(x-\beta) = x^2 - (\alpha + \beta) x + \alpha \beta.$$ Equating coefficients in $x$, we get $\alpha + \beta = 1$, $\alpha \beta = -1$. Thus $$P_{k+1} = P_k + P_{k-1}.$$ Now since $$P_0 = \alpha^0 + \beta^0 = 2,$$ we compute the recursion for $P_k$ easily: $$P_1 = 1, \\ P_2 = P_1 + P_0 = 3, \\ P_3 = P_2 + P_1 = 4, \\ \text{etc.}$$