Q. 21 - Calculus

Question

The Fibonacci numbers may be computed with the formula

$f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}}$

Use this formula to compute f$f_1, f_2, f_3, f_4, \mathrm{and} f_5.$(Imagine computing $f_{100}$.)

Step-by-Step Solution

Answer

The first five terms of the sequence are $f_1=1, f_2=1, f_3=2, f_4=3, \& f_5=5.$

Step 1. Given information.

The given sequence is the following.

$f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}}$

Step 2. The first term of the sequence.

Substitute $k=1$in the sequence for the first term.

$$\begin{gathered} f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}} \\ {f}_1=\frac{{\left(1+\sqrt{5}\right)}^1-{\left(1-\sqrt{5}\right)}^1}{2^1\sqrt{5}} \\ {f}_1=\frac{2\sqrt{5}}{2\sqrt{5}} \\ {f}_1=1 \end{gathered}$$

So the first term of the sequence is $f_1=1.$

Step 3. The second term of the sequence.

Substitute $k=2$in the sequence for the second term.

$$\begin{gathered} f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}} \\ {f}_2=\frac{{\left(1+\sqrt{5}\right)}^2-{\left(1-\sqrt{5}\right)}^2}{2^2\sqrt{5}} \\ {f}_2=\frac{4\sqrt{5}}{4\sqrt{5}} \\ {f}_2=1 \end{gathered}$$

So the second term of the sequence is $f_2=1.$

Step 4. The third term of the sequence.

Substitute $k=3$in the sequence for the third term.

$$\begin{gathered} f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}} \\ {f}_3=\frac{{\left(1+\sqrt{5}\right)}^3-{\left(1-\sqrt{5}\right)}^3}{2^3\sqrt{5}} \\ {f}_3=\frac{16\sqrt{5}}{8\sqrt{5}} \\ {f}_3=2 \end{gathered}$$

So the third term of the sequence is $f_3=2.$

Step 5. The fourth term of the sequence.

Substitute $k=4$in the sequence for the fourth term.

$$\begin{gathered} f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}} \\ {f}_4=\frac{{\left(1+\sqrt{5}\right)}^4-{\left(1-\sqrt{5}\right)}^4}{2^4\sqrt{5}} \\ {f}_4=\frac{48\sqrt{5}}{16\sqrt{5}} \\ {f}_4=3 \end{gathered}$$

So the fourth term of the sequence is $f_4=3.$

Step 6. The fifth term of the sequence.

Substitute $k=5$in the sequence for the fifth term.

$$\begin{gathered} f_k=\frac{{\left(1+\sqrt{5}\right)}^k-{\left(1-\sqrt{5}\right)}^k}{2^k\sqrt{5}} \\ {f}_5=\frac{{\left(1+\sqrt{5}\right)}^5-{\left(1-\sqrt{5}\right)}^5}{2^5\sqrt{5}} \\ {f}_5=\frac{160\sqrt{5}}{32\sqrt{5}} \\ {f}_5=5 \end{gathered}$$

So the fifth term of the sequence is $f_5=5.$