Question

If $x^2-5x+1=0$, then $x^5+\frac{1}{x^5}$ is equal to

• A. $3125$
• B. $2625$
• C. $3025$
• D. $2525$

Answer 1

Answer: (D)

$2525$

Let roots of given equation be $\alpha$ and $\beta$

$\alpha .\beta =1$

$\Rightarrow \beta =\frac{1}{\alpha }$

Hence ${\alpha }^5+(\frac{1}{\alpha }{)}^5$ is same as ${\alpha }^5+{\beta }^5$

Now

${\alpha }^2-5\alpha +1=0$

${\beta }^2-5\beta +1=0$

multiplying by ${\alpha }^{n-2}$ and ${\beta }^{n-2}$ and reaarranging

${\alpha }^n=5{\alpha }^{n-1}-{\alpha }^{n-2}$

${\beta }^n=5{\beta }^{n-1}-{\beta }^{n-2}$

Adding we get

${\alpha }^n+{\beta }^n=5{\alpha }^{n-1}+5{\beta }^{n-1}-{\alpha }^{n-2}-{\beta }^{n-2}$

let ${\alpha }^n+{\beta }^n$ be $A_n$

Then

$A_n=5A_{n-1}-A_{n-2}$

$A_0=2$ and

$A_1=5$ sum of roots.

Hence

$A_2=25-2=23$

$A_3=115-5=110$

$A_4=550-23=527$

$A_5=2635-110=2525$