The World of Numbers | EOT

Question 15

Show that the rational number (a+b)2\frac{(a+b)}{2} lies between the rational numbers aa and bb.

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Solution

We will use the given recurrence relation to find each term step by step.

Step 1 — Calculate W3W_3

We are given the values for W1W_1 and W2W_2. The formula for n>2n > 2 is Wn=W1+W2++Wn2+2W_n = W_1 + W_2 + \dots + W_{n-2} + 2. Let's find W3W_3.

W3=W1+2W_3 = W_1 + 2

W3=1+2W_3 = 1 + 2

W3=3\boxed{W_3 = 3}

Step 2 — Calculate W4W_4

We use the recurrence relation again. Let's find W4W_4.

W4=W1+W2+2W_4 = W_1 + W_2 + 2

W4=1+2+2W_4 = 1 + 2 + 2

W4=5\boxed{W_4 = 5}

Step 3 — Calculate W5W_5

We continue finding the terms. Let's find W5W_5.

W5=W1+W2+W3+2W_5 = W_1 + W_2 + W_3 + 2

W5=1+2+3+2W_5 = 1 + 2 + 3 + 2

W5=8\boxed{W_5 = 8}

Step 4 — Calculate W6W_6

Let's find the next term, W6W_6.

W6=W1+W2+W3+W4+2W_6 = W_1 + W_2 + W_3 + W_4 + 2

W6=1+2+3+5+2W_6 = 1 + 2 + 3 + 5 + 2

W6=13\boxed{W_6 = 13}

Step 5 — Calculate W7W_7

We find the value for W7W_7.

W7=W1+W2+W3+W4+W5+2W_7 = W_1 + W_2 + W_3 + W_4 + W_5 + 2

W7=1+2+3+5+8+2W_7 = 1 + 2 + 3 + 5 + 8 + 2

W7=21\boxed{W_7 = 21}

Step 6 — Calculate W8W_8

Finally, we find the value for W8W_8.

W8=W1+W2+W3+W4+W5+W6+2W_8 = W_1 + W_2 + W_3 + W_4 + W_5 + W_6 + 2

W8=1+2+3+5+8+13+2W_8 = 1 + 2 + 3 + 5 + 8 + 13 + 2

W8=34\boxed{W_8 = 34}

Step 7 — List the sequence and identify it

We have found all the required terms. The sequence starts with W1=1W_1 = \mathbf{1} and W2=2W_2 = \mathbf{2}. The calculated terms are W3=3W_3 = \mathbf{3}, W4=5W_4 = \mathbf{5}, W5=8W_5 = \mathbf{8}, W6=13W_6 = \mathbf{13}, W7=21W_7 = \mathbf{21}, W8=34W_8 = \mathbf{34}. This sequence follows a specific pattern. Each term is the sum of the previous two terms. For example, W3=W2+W1=2+1=3W_3 = W_2 + W_1 = 2 + 1 = 3. This is a well-known sequence.

Diagram 1

Answer

(i) The values of W1,W2,,W8W_1, W_2, \dots, W_8 are 1, 2, 3, 5, 8, 13, 21, 34. (ii) Yes, this is the Virahanka-Fibonacci sequence.

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