Question

If $z^2-2z+4=0$ then the value of ${\left(\frac{z}{2}+\frac{2}{z}\right)}^2+{\left(\frac{z^2}{4}+\frac{4}{z^2}\right)}^2+{\left(\frac{z^3}{8}+\frac{8}{z^3}\right)}^2+........+{\left(\frac{z^{24}}{2^{24}}+\frac{2^{24}}{Z^{24}}\right)}^2$ is equal to.

• A. $24$
• B. $32$
• C. $48$
• D. None of these

Answer 1

The correct option is A $24$

Given :- $z^2-2z+4=0$

$z^2+4=2x$

Problem : ${\sum }_{n=1}^{24}{(\frac{z^n}{2^n}+\frac{2^n}{z^n})}^2$

${\sum }_{n=1}^{24}{\left(\frac{z^{2n}+2^{2n}}{(2z{)}^n}\right)}^2$

We know $(z^{2^{\left(1\right)}}+2^{2^{\left(2\right)}})=2z^{\left(1\right)}$

$(z^2+2^2)=(2z{)}^2$

$z^4+2^4+2.z^2{2}^2=(2z{)}^2$

$z^4+2^4=2^4{z}^2-8z^2$

$8z^2$

When $n=1{\left(\frac{z^2+2^2}{2z}\right)}^2$

${\left(\frac{2z}{2z}\right)}^2$

$=1.$

When $n=2\frac{(z^4+2^4{)}^2}{(2^2{z}^2{)}^2}$

$\frac{z^4+2^4+2.z^2{.2}^2}{(2^2{z}^2{)}^2}$

$\frac{8z^2}{8z^2}$

$=1$.

So on

$\therefore {\sum }_{n=1}^{24}{\left(\frac{z^{2n}+2^{2n}}{(2z{)}^n}\right)}^2=\sum {\left(\frac{(2z{)}^n}{(2z{)}^n}\right)}^2$

$={\sum }_{n=1}^{24}1$.

$=24$