If the first three Fibonacci numbers are x1 = 1, x2 = 1, x3 = 2, what is the value of n for which xn + xn+1 + xn+2= 68?

13, 21, and 34

Step-by-step explanation:

x(n) + x(n+1) + x(n+2)= 68

x(n), x(n+1), and x(n+2) are three consecutive Fibonacci numbers.

=> (x) + (x+k) + (x+x+k)=68

(x is a Fibonacci number and k is a positive integer)

=> 4x + 2k = 68

=> 2x + k = 34

=> x < 17

Check two Fibonacci numbers that are closest to 17 and smaller than 17.

x(n)=8 => x(n+1)=13, x(n+2)= 21

=>x(n) + x(n+1) + x(n+2)= 8 + 13 + 21 = 42 (not valid)

x(n) = 13 => x(n+1) = 21, x(n+2)= 34

=> x(n) + x(n+1) + x(n+2)= 13 + 21 + 34 = 68 (valid)

Hope this helps!

i think the answer is true

step-by-step explanation:

d no solution

step-by-step explanation: