Question

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

Answer 1

`11`.

Explanation:

Given that, `a-1/a=1`, on squaring, we have,

`a^2-2*a*1/a+1/a^2=1, or,`

`a^2+1/a^2=3..................................................................(square_1)`.

Also, `a^3-1/a^3=(a-1/a)(a^2+1+1/a^2)`,

`=(1)(3+1)....................................................[because, (square_1)]`.

`rArr a^3-1/a^3=4.........................................................(square_2)`.

Utilising `(square_1) and (square_2)`, we have,

`(a^2+1/a^2)(a^3-1/a^3)=3xx4=12`.

`:. a^5-a^2/a^3+a^3/a^2-1/a^5=12, i.e.,`

`a^5-1/a+a-1/a^5=12`.

`rArr a^5-1/a^5=12-a+1/a=12-(1)=11`.

Answer 2

11

Explanation:

From the binomial theorem, we know that

`(a+b)^3 = a^3+3a^2b+3ab^2+b^3`

and

`(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5`

Thus

`(a-1/a)^3 = a^3-3a^2 1/a+3a(1/a)^2-(1/a)^3`
`qquad = (a^3-1/a^3)-3(a-1/a)`

So, `a-1/a=1 implies`
`a^3-1/a^3=(a-1/a)^3+3(a-1/a)=1^3+3times 1=4`

Similarly, we show that

`(a-1/a)^5=color(red)(a^5)+color(blue)(5a^4(-1/a))+10a^3(-1/a)^2+10a^2(-1/a)^3+color(blue)(5a(-1/a)^4)+color(red)((-1/a)^5)`
`=color(red)(a^5)color(blue)(-5a^3)+10a-10/a color(blue)(+5/a^3) color(red)(-1/a^5)`
`= color(red)((a^5-1/a^5))color(blue)(-5(a^3-1/a^3))+10(a-1/a)`

Thus

`a^5-1/a^5 = (a-1/a)^5 + 5(a^3-1/a^3)-10(a-1/a)`

so that, for our problem

`a^5-1/a^5=1^5+5times 4-10times 1=11`

Answer 3

`a^5-1/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-1/a = 1`

Then:

`a^2 = a + 1`

`1/a^2 = 1-1/a`

`a^2+1/a^2 = (a-1/a)^2+2 = 1+2 = 3`

Consider the sequence with general term:

`L_n = a^n+(-1)^n/a^n`

We find:

`L_(2k+2) = a^(2k+2)+1/a^(2k+2)`

`color(white)(L_(2k+2)) = a^2 a^(2k) + 1/a^2 1/a^(2k)`

`color(white)(L_(2k+2)) = (a+1) a^(2k) + (1-1/a) 1/a^(2k)`

`color(white)(L_(2k+2)) = a^(2k+1)-1/a^(2k+1)+a^(2k) + 1/a^(2k)`

`color(white)(L_(2k+2)) = L_(2k+1)+L_(2k)`

Also:

`L_(2k+1) = a^(2k+1)-1/a^(2k+1)`

`color(white)(L_(2k+1)) = a^2 a^(2k-1) - 1/a^2 1/a^(2k-1)`

`color(white)(L_(2k+1)) = (a+1) a^(2k-1) - (1-1/a) 1/a^(2k-1)`

`color(white)(L_(2k+1)) = a^(2k)+1/a^(2k)+a^(2k-1)-1/a^(2k-1)`

`color(white)(L_(2k+1)) = L_(2k)+L_(2k-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+1/a^0 = 1+1/1 = 2`

`L_1 = a^1-1/a^1 = a-1/a = 1`

So we have a complete recursive definition of `L_n`

`{ (L_0 = 2), (L_1 = 1), (L_(n+2) = L_(n+1)+L_n) :}`

This sequence starts:

`2, 1, 3, 4, 7, 11`

So:

`a^5-1/a^5 = L_5 = 11`

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