The difference between each successive term is beginning to decrease and is almost at zero.
→As ‘n’ keeps increasing, the value of an – an+1 keeps decreasing , and gradually at that.
The exact value of this infinite surd is 1.62.
Q3) Consider another infinite surd, where the first term is Repeat the entire process above and find the exact value for this surd.
Answer) Basically repeat the same process for the first infinite surd that had 1, instead of 2.
So, in this case we take the surd as a sequence of terms, where is;
a1 =
so,
a2 =
a3 =
also,
a2 =
a3 =
Squaring on both sides
(a3)2 = ()2
(a3)2 = 2 + a2
a2 = (a3)2 – 2
an = (an+1)2 – 2
Therefore,
an+1 =
Again, we have to find the terms’ exact values.
So,
-
n Term (an) an – an-1
- 1 1.847759 1.847759
- 2 1.961571 0.113811
- 3 1.990369 0.028799
- 4 1.997591 0.007221
- 5 1.999398 0.001807
- 6 1.999849 0.000452
- 7 1.999962 0.000113
-
8 1.999991 2.82 x 10-5
-
9 1.999998 7.06 x 10-6
-
10 1.999999 1.76 x 10-6
And now a graph of the relation between n and the nth term:
The difference between each successive term is decreasing, and eventually it will reach zero.
Since ‘n’ keeps increasing, the value of - keeps decreasing, gradually.
The exact value of this surd is 2.
Q4) Now consider the general infinite surd where the first term is . Find an expression for the exact value of this general infinite surd in terms of ‘k’.
Answer)
In the general infinite surd:
a1 =
a2 =
also,
a2 =
a3 =
Squaring both sides,
(a3)2 = ()2
(a3)2 = a2 + k
a2 = (a3)2 – k
Therefore,
an-1 = (an)2 – k
so, k = (an)2 – an-1
Q5) Find some value of ‘k’ that makes the expression an integer. Find the general statement that represents all the values of ‘k’ for which the expression is an integer.
- I used Apple iWork’09 Numbers to find the values of k that are integers.
I used the formula;
a1 =
a2 =
a3 =
and so on.
Therefore,
When k = 2, the equation gets a value as an integer at a12When k = 6, the equation gets a value as an integer at a9
So,
When k = 12, the equation gets a value as an integer at a8
When k = 20, the equation gets a value as an integer at a8
When k = 30, the equation gets a value as an integer at a7
The values of ‘k’ that make the expression an integer are:
2, 6, 12, 20, 30...
Based on all this, I got to my general statement, and it is;
an = 2 (a(n-1) + 1) – an-2
Q6) Test the validity of your statement by using different values of ‘k’.
Answer)
a5 = 2 (a(5-1) + 1) – a(5-2)
= 2 (a4 + 1) – a3
= 2 (20 + 1) – 12
= 42 – 12
= 30
a4 = 2 (a(4-1) + 1) – a(4-2)
= 2 (a3 + 1) – a2
= 2 (12 + 1) - 6
= 26 – 6
= 20
So, the general statement that we concluded is completely valid.
Q7) Discuss the scope and/or limitation of your general statement.
Answer) To know any of these terms, we need to know about its 2 preceding terms, and because of this, we may not be able to find the first 2 terms. They have to be given, as there are no terms that precede the first 2 terms. I only found values up to the 5th term; therefore I drew this conclusion from my analysis of the behaviour of those terms.
Q8) Explain how you arrived at your general statement.
Answer)
As is said above, the successive term can be obtained by the relation;
an = 2 (an-1 + 1) – an-2
