Question

Let $a,b$ be the roots of the quadratic equation $x^2+px+1=0,$ where $p^2=1$.
If $k$ is a positive integer, then $\sum _{n=1}^k(a^{2^n}+b^{2^n})$ is

• A. $0$
• B. $k$
• C. $-k$
• D. $0$ if $k$ is even and $-1$ if $k$ is odd.

Answer 1

The correct option is C $-k$

By the given quadratic equation, $ab=1$ and $a+b=\pm 1$
$\sum _{n=1}^k(a^{2^n}+b^{2^n})=a^2+b^2+a^4+b^4+\cdots +(a^{2^k}+b^{2^k})$

Now, $a^2+b^2=(a+b{)}^2-2ab=1-2=-1$
$a^4+b^4=(a^2+b^2{)}^2-2a^2{b}^2=1-2=-1$
and so on.
Each term $(a^{2^n}+b^{2^n})$ in that summation is equal to $-1$.
Hence, $\sum _{n=1}^k(a^{2^n}+b^{2^n})=-k$