JOMO 7, Short 2

Given $w^2-1=w$ $=>(w-\frac{1}{w})=1$ Squaring, we get $(w^2+\frac{1}{w^2})=3$ Now, $(w^2+\frac{1}{w^2})^5=(w^{10}+\frac{1}{w^{10}})+5(w^6+\frac{1}{w^6})+10(w^2+\frac{1}{w^2})$ Now we need $(w^6+\frac{1}{w^6})$ We get this term here, $(w^2+1/w^2)^3=(w^6+\frac{1}{w^6})+3(w^2+1/w^2)$ On solving, $(w^6+\frac{1}{w^6})=18$ Putting all the values in our main equation, $(w^{10}+1/w^{10})=3^5-5(18)-10(3)=\boxed{123}$

When expanding $(w-1)(w+1)=w$ we get $w^2-w-1 =0$ , which is the characteristic equation of any Fibonnaci-like sequence, given by $F_n=F_{n-1}+F_{n-2}$ . General solutions to this recurrence relation are of the form $F_n=aw^n+b\frac{1}{w^n}$ for $w=\frac{1+\sqrt{5}}{2}$ , since $w$ and $\frac{1}{w}$ are the zeroes of the equation. The values for $a$ and $b$ relate to $F_0$ and $F_1$ .

In our current case, we want to know $w^{10}+\frac{1}{w^{10}}$ , so we want to know the tenth value in the sequence corresponding to $a=b=1$ . We find that $F_0=2, F_1=1$ and easily calculate the successive terms of $(F_n)$ from here: $3, 4, 7, 11, 18, 29, 47, 76, 123$ and so $F_n=\boxed{123}$ .

solution

$(w-1)(w+1)=w$

$w^{2}-1=w$

Dividing both sides by $w$ ,we get

$w-\frac{1}{w}=1$

Squaring and adding $2$ to both sides,we have

$\boxed{w^{2}+\frac{1}{w^{2}}=3}$

Simultaneously squaring and subtracting $2$ from both sides,we get the following equations:

$\boxed{w^{4}+\frac{1}{w^{4}}=7}$ and $\boxed{w^{8}+\frac{1}{w^{8}}=47}$

Now note that: $w^{10}+\frac{1}{w^{10}}=(w^{2}+\frac{1}{w^{2}})$ $(w^{8}+\frac{1}{w^{8}})$ - $(w^{6}+\frac{1}{w^{6}})$

In this equation we only have to find the value of $(w^{6}+\frac{1}{w^{6}})$

Now $(w^{6}+\frac{1}{w^{6}})$

= $(w^{2}+\frac{1}{w^{2}})^{3}-3(w^{2}+\frac{1}{w^{2}})$

= $3^{3}-3.3$

= $18$

Now substituting this value in the main equation,

$w^{10}+\frac{1}{w^{10}}$

= $(w^{2}+\frac{1}{w^{2}})$ $(w^{8}+\frac{1}{w^{8}})$ - $18$

= $3.47-18$

= $\boxed{123}$

w+1/w=sqrt(5) as

By the first equation, $w = \pm \phi = \frac{\sqrt{5} \pm 1}{2}$ .

Then $w^{10} + \frac{1}{w^{10}} = \frac{1}{2^{10}} \left(\sum_{i=0}^{10} \left( {10 \choose i} \sqrt{5}^i (-1)^{10-i} \right) + \sum_{i=0}^{10} \left( {10 \choose i} \sqrt{5}^i \right) \right)= \frac{1}{2^9} \sum_{k=0}{5} \left( {10 \choose 2k} \sqrt{5}^{2k} \right) = \frac{1}{2^9} 62976$

(w^2)-1=w or w-(1/w)=1 or (w-1/w)^2=1^2 [squaring both sides] or w^2+(1/w^2)-2=1 or w^2+(1/w^2)=1+2 or (w+1/w)^2-2=3 or (w+1/w)^2=5 so w+1/w=sqrt 5,Now w^10+(1/w)^10=(w^5)^2+(1/w^5)^2=(w^5+1/w^5)^2-2={(w^3+1/w^3)(w^2+1/w^2)-(w+1/w)}^2-2=[{(w+1/w)^3-3(w+1/w)}{(w+1/w)^2-2}-(w+1/w)]^2-2=[{(sqrt 5)^3-3 sqrt 5}{(sqrt 5)^2-2}-sqrt 5}^2-2=[(5 sqrt5-3 sqrt5) 3-(sqrt 3)]^2-2=[2 sqrt5 3-sqrt5]^2-2=[6 sqrt5-sqrt5]^2-2=[5 sqrt5]^2-2=125-2=123(ans)