Question

$\bf \: if \: a + \frac{1}{a} = 2 \: \: then \: \: find \: the \: \\ \bf \: value \: of \: \: {a}^{2004} + \frac{1}{ {a}^{2004} }$ Very urgent...​+ challenge

Answer 1

Answer :

${a}^{2004} + \dfrac{1}{ {a}^{2004} } = 2$

Explanation :

We are given,

$\implies a + \dfrac{1}{a} = 2$

$\implies \dfrac{ {a}^{2} + 1}{a} = 2$

$\implies {a}^{2} + 1 = 2a$

$\implies {a}^{2} - 2a + 1 = 0$

Now the only solution of this equation is 1.

Hence the value of a is 1.

Now,

$\implies{a}^{2004} + \dfrac{1}{ {a}^{2004} }$

Substitute a = 1

$\implies{1}^{2004} + \dfrac{1}{ {1}^{2004} }$

$\implies \: 1 + 1$

$\implies \: 2$

Hence solved.

Answer 2

Step-by-step explanation:

We are given,

\implies a + \dfrac{1}{a} = 2⟹a+

a

1

=2

\implies \dfrac{ {a}^{2} + 1}{a} = 2⟹

a

a

2

+1

=2

\implies {a}^{2} + 1 = 2a⟹a

2

+1=2a

\implies {a}^{2} - 2a + 1 = 0⟹a

2

−2a+1=0

Now the only solution of this equation is 1.

Hence the value of a is 1.

Now,

\implies{a}^{2004} + \dfrac{1}{ {a}^{2004} }⟹a

2004

+

a

2004

1

Substitute a = 1

\implies{1}^{2004} + \dfrac{1}{ {1}^{2004} }⟹1

2004

+

1

2004

1

\implies \: 1 + 1⟹1+1

\implies \: 2⟹2

Hence solved.