When #a-1/a=1#, #a^5-1/a^5 =#? how to solve....
3 Answers
Explanation:
Given that,
Also,
Utilising
11
Explanation:
From the binomial theorem, we know that
and
Thus
So,
Similarly, we show that
Thus
so that, for our problem
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
#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_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.