# When a-1/a=1, a^5-1/a^5 =? how to solve....

Mar 21, 2018

$11$.

#### Explanation:

Given that, $a - \frac{1}{a} = 1$, on squaring, we have,

${a}^{2} - 2 \cdot a \cdot \frac{1}{a} + \frac{1}{a} ^ 2 = 1 , \mathmr{and} ,$

${a}^{2} + \frac{1}{a} ^ 2 = 3. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\square}_{1}\right)$.

Also, ${a}^{3} - \frac{1}{a} ^ 3 = \left(a - \frac{1}{a}\right) \left({a}^{2} + 1 + \frac{1}{a} ^ 2\right)$,

$= \left(1\right) \left(3 + 1\right) \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left[\because , \left({\square}_{1}\right)\right]$.

$\Rightarrow {a}^{3} - \frac{1}{a} ^ 3 = 4. \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left({\square}_{2}\right)$.

Utilising $\left({\square}_{1}\right) \mathmr{and} \left({\square}_{2}\right)$, we have,

$\left({a}^{2} + \frac{1}{a} ^ 2\right) \left({a}^{3} - \frac{1}{a} ^ 3\right) = 3 \times 4 = 12$.

$\therefore {a}^{5} - {a}^{2} / {a}^{3} + {a}^{3} / {a}^{2} - \frac{1}{a} ^ 5 = 12 , i . e . ,$

${a}^{5} - \frac{1}{a} + a - \frac{1}{a} ^ 5 = 12$.

$\Rightarrow {a}^{5} - \frac{1}{a} ^ 5 = 12 - a + \frac{1}{a} = 12 - \left(1\right) = 11$.

Mar 21, 2018

11

#### Explanation:

From the binomial theorem, we know that

${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$

and

${\left(a + b\right)}^{5} = {a}^{5} + 5 {a}^{4} b + 10 {a}^{3} {b}^{2} + 10 {a}^{2} {b}^{3} + 5 a {b}^{4} + {b}^{5}$

Thus

${\left(a - \frac{1}{a}\right)}^{3} = {a}^{3} - 3 {a}^{2} \frac{1}{a} + 3 a {\left(\frac{1}{a}\right)}^{2} - {\left(\frac{1}{a}\right)}^{3}$
$q \quad = \left({a}^{3} - \frac{1}{a} ^ 3\right) - 3 \left(a - \frac{1}{a}\right)$

So, $a - \frac{1}{a} = 1 \implies$
${a}^{3} - \frac{1}{a} ^ 3 = {\left(a - \frac{1}{a}\right)}^{3} + 3 \left(a - \frac{1}{a}\right) = {1}^{3} + 3 \times 1 = 4$

Similarly, we show that

${\left(a - \frac{1}{a}\right)}^{5} = \textcolor{red}{{a}^{5}} + \textcolor{b l u e}{5 {a}^{4} \left(- \frac{1}{a}\right)} + 10 {a}^{3} {\left(- \frac{1}{a}\right)}^{2} + 10 {a}^{2} {\left(- \frac{1}{a}\right)}^{3} + \textcolor{b l u e}{5 a {\left(- \frac{1}{a}\right)}^{4}} + \textcolor{red}{{\left(- \frac{1}{a}\right)}^{5}}$
$= \textcolor{red}{{a}^{5}} \textcolor{b l u e}{- 5 {a}^{3}} + 10 a - \frac{10}{a} \textcolor{b l u e}{+ \frac{5}{a} ^ 3} \textcolor{red}{- \frac{1}{a} ^ 5}$
$= \textcolor{red}{\left({a}^{5} - \frac{1}{a} ^ 5\right)} \textcolor{b l u e}{- 5 \left({a}^{3} - \frac{1}{a} ^ 3\right)} + 10 \left(a - \frac{1}{a}\right)$

Thus

${a}^{5} - \frac{1}{a} ^ 5 = {\left(a - \frac{1}{a}\right)}^{5} + 5 \left({a}^{3} - \frac{1}{a} ^ 3\right) - 10 \left(a - \frac{1}{a}\right)$

so that, for our problem

${a}^{5} - \frac{1}{a} ^ 5 = {1}^{5} + 5 \times 4 - 10 \times 1 = 11$

Mar 24, 2018

${a}^{5} - \frac{1}{a} ^ 5 = {L}_{5} = 11$

#### Explanation:

One interesting thing about this question is that it relates to a sequence defined by a recursive rule, similar to a Fibonacci sequence.

Given:

$a - \frac{1}{a} = 1$

Then:

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

$\frac{1}{a} ^ 2 = 1 - \frac{1}{a}$

${a}^{2} + \frac{1}{a} ^ 2 = {\left(a - \frac{1}{a}\right)}^{2} + 2 = 1 + 2 = 3$

Consider the sequence with general term:

${L}_{n} = {a}^{n} + {\left(- 1\right)}^{n} / {a}^{n}$

We find:

${L}_{2 k + 2} = {a}^{2 k + 2} + \frac{1}{a} ^ \left(2 k + 2\right)$

$\textcolor{w h i t e}{{L}_{2 k + 2}} = {a}^{2} {a}^{2 k} + \frac{1}{a} ^ 2 \frac{1}{a} ^ \left(2 k\right)$

$\textcolor{w h i t e}{{L}_{2 k + 2}} = \left(a + 1\right) {a}^{2 k} + \left(1 - \frac{1}{a}\right) \frac{1}{a} ^ \left(2 k\right)$

$\textcolor{w h i t e}{{L}_{2 k + 2}} = {a}^{2 k + 1} - \frac{1}{a} ^ \left(2 k + 1\right) + {a}^{2 k} + \frac{1}{a} ^ \left(2 k\right)$

$\textcolor{w h i t e}{{L}_{2 k + 2}} = {L}_{2 k + 1} + {L}_{2 k}$

Also:

${L}_{2 k + 1} = {a}^{2 k + 1} - \frac{1}{a} ^ \left(2 k + 1\right)$

$\textcolor{w h i t e}{{L}_{2 k + 1}} = {a}^{2} {a}^{2 k - 1} - \frac{1}{a} ^ 2 \frac{1}{a} ^ \left(2 k - 1\right)$

$\textcolor{w h i t e}{{L}_{2 k + 1}} = \left(a + 1\right) {a}^{2 k - 1} - \left(1 - \frac{1}{a}\right) \frac{1}{a} ^ \left(2 k - 1\right)$

$\textcolor{w h i t e}{{L}_{2 k + 1}} = {a}^{2 k} + \frac{1}{a} ^ \left(2 k\right) + {a}^{2 k - 1} - \frac{1}{a} ^ \left(2 k - 1\right)$

$\textcolor{w h i t e}{{L}_{2 k + 1}} = {L}_{2 k} + {L}_{2 k - 1}$

So for any $n$, odd or even, we have:

${L}_{n + 2} = {L}_{n + 1} + {L}_{n}$

For our initial terms we find:

${L}_{0} = {a}^{0} + \frac{1}{a} ^ 0 = 1 + \frac{1}{1} = 2$

${L}_{1} = {a}^{1} - \frac{1}{a} ^ 1 = a - \frac{1}{a} = 1$

So we have a complete recursive definition of ${L}_{n}$

$\left\{\begin{matrix}{L}_{0} = 2 \\ {L}_{1} = 1 \\ {L}_{n + 2} = {L}_{n + 1} + {L}_{n}\end{matrix}\right.$

This sequence starts:

$2 , 1 , 3 , 4 , 7 , 11$

So:

${a}^{5} - \frac{1}{a} ^ 5 = {L}_{5} = 11$

I used the letter "L" intentionally, as this is known as the Lucas sequence.