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STEP III 2007 question 3 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 2007 question 3 solution

(i)

F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, F_5 = 5, F_6 = 8, F_7 = 13, F_8 = 21. Fibonacci, yay!

(ii)

F_{2k+3}F_{2k+1} - F_{2k+2}^2

= (F_{2k+2}+F_{2k+1})F_{2k+1} - F_{2k+2}^2

= F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}^2

= F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}(F_{2k+1}F_{2k})

= F_{2k+2}F_{2k+1} +F_{2k+1}^2 - F_{2k+2}F_{2k+1} - F_{2k+2}F_{2k}

= - F_{2k+2}F_{2k}+F_{2k+1}^2.

(iii)

P(n): \quad F_{2n+1}F_{2n-1} - F_{2n}^2 = 1

Choosing n=2 gives:

\text{LHS} = F_5F_3 - F_4^2 = 5\cdot 2 - 3^2 = 10 - 9 = 1 = \text{RHS} .

Assume P(k):

P(k) \Rightarrow F_{2k+1}F_{2k-1} - F_{2k}^2 = 1

\Rightarrow F_{2k+1}(F_{2k+1}-F_{2k}) - F_{2k}^2 = 1

\Rightarrow F_{2k+1}(F_{2k+1}-F_{2k}) - F_{2k}(F_{2k+2}-F_{2k+1}) = 1

\Rightarrow F_{2k+1}^2 - F_{2k+1}F_{2k} - F_{2k}F_{2k+2} + F_{2k}F_{2k+1} = 1

\Rightarrow F_{2k+1}^2 - F_{2k}F_{2k+2} = 1

\Rightarrow F_{2k+1}(F_{2k+3}-F_{2k+2}) - (F_{2k+2}-F_{2k+1})F_{2k+2} = 1

\Rightarrow F_{2k+1}F_{2k+3} - F_{2k+1}F_{2k+2} - F_{2k+2}^2+F_{2k+1}F_{2k+2} = 1

\Rightarrow F_{2k+1}F_{2k+3} - F_{2k+2}^2 = 1

\Rightarrow F_{2(k+1)+1}F_{2(k+1)-1} - F_{2(k+1)}^2 = 1 \Rightarrow P(k+1).

P(k) \Rightarrow P(k+1) which completes the induction.

Then \displaystyle \frac{F_{2n}^2 + 1}{F_{2n+1}} = F_{2n-1} which is an integer.

(iv)

F_{2n-1}^2 + 1

= (F_{2n+1} - F_{2n})^2 + 1

= (F_{2n+1} - F_{2n})^2 + (F_{2n+1}F_{2n-1} - F_{2n}^2)

= F_{2n+1}^2 - 2F_{2n+1}F_{2n} + F_{2n}^2 + F_{2n+1}F_{2n-1} - F_{2n}^2

= F_{2n+1}(F_{2n+1} - 2F_{2n} + F_{2n-1}) as required.

Solution by generalebriety.

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