Answer:
The value of n for which xₙ₋₂ + xₙ₋₁ + xₙ = 68 is n = 9
Step-by-step explanation:
The nth term of Fibonacci number is given by the following relation;
[tex]F_n = \dfrac{(1 + \sqrt{5} )^n - (1 - \sqrt{5} )^n }{2^n\sqrt{5} }[/tex]
Given that we have ;
[tex]\Sigma x_n = \dfrac{(1 + \sqrt{5} )^{n-2} - (1 - \sqrt{5} )^{n-2} }{2^{n-2}\sqrt{5} } + \dfrac{(1 + \sqrt{5} )^{n-1} - (1 - \sqrt{5} )^{n-1} }{2^{n-1}\sqrt{5} }+\dfrac{(1 + \sqrt{5} )^{n} - (1 - \sqrt{5} )^{n} }{2^{n}\sqrt{5} }[/tex]When n = 3, we have;
∑xₙ = 4
When n = 4 we have
∑xₙ = 4
When n = 6 we have
∑xₙ = 16
When n = 9 we have
∑xₙ = 68
Therefore, we have the value of n for which xₙ₋₂ + xₙ₋₁ + xₙ = 68 is n = 9.
