Question

if $x^2-x+1=0$ then the value of $\sum _{n=1}^5{[x^n+\frac{1}{x^n}]}^2$

• A. $8$
• B. $10$
• C. $12$
• D. None of these

Answer 1

Answer: (A)

$8$

$x^2-x+1=0$
Multiplying both sides by $x+1$ gives $x^3=-1$
The expression we need is $\sum _{n=1}^5{[x^n+\frac{1}{x^n}]}^2$
The first term is $[x+\frac{1}{x}{]}^2=1$
The second term is $[x^2+\frac{1}{x^2}{]}^2=[\frac{x^4+1}{x^2}{]}^2=[\frac{-x+1}{x^2}{]}^2=1$ ...(Using $x^3=-1$)
The third term is $[x^3+\frac{1}{x^3}{]}^2=[-1-1{]}^2=4$
The fourth term is $[x^4+\frac{1}{x^4}{]}^2=[-x+\frac{-1}{x}{]}^2=1$
The fifth term is $[x^5+\frac{1}{x^5}{]}^2=[-x^2+\frac{-1}{x^2}{]}^2=1$
Adding all, we have the value of the expression to be $1+1+4+1+1=8.$