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Route to Root

A sequence of numbers x1, x2, x3, ... starts with x1 = 2, and, if you know any term xn, you can find the next term xn+1 using the formula: xn+1 = (xn + 3/xn)/2 . Calculate the first six terms of this sequence. What do you notice? Calculate a few more terms and find the squares of the terms. Can you prove that the special property you notice about this sequence will apply to all the later terms of the sequence? Write down a formula to give an approximation to the cube root of a number and test it for the cube root of 3 and the cube root of 8. How many terms of the sequence do you have to take before you get the cube root of 8 correct to as many decimal places as your calculator will give? What happens when you try this method for fourth roots or fifth roots etc.?

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Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

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LOGO Challenge - Sequences and Pentagrams

Explore this how this program produces the sequences it does. What are you controlling when you change the values of the variables?

Weekly Challenge 34: Googol

Stage: 5 Short Challenge Level: Challenge Level:2 Challenge Level:2

The numbers in the problem are too large for computers to deal with in a straightforward fashion (or, rather, were in 2011), so we need to use pure mathematics to help us. There are at least two possible positive ways forwards:
 
First, you might notice that the inequality is a quadratic in the variable $n^2$. You could solve the corresponding equality and use this to work out the minimum value of $n$ by rounding up the answer to the next largest integer.
 
Alternatively, you could notice that $10^{100}=(10^{25})^4$. So, it is quite clear that $n=10^{25}$ is too small.
 
What about $10^{25}+1$? We can substitute this value and use the binomial theorem to show that $$ \begin{eqnarray} (10^{25}+1)^4-6(10^{25}+1)^2&=&\left(10^{100} +4\times 10^{75} + 6\times 10^{50} +4\times 10^{25} + 1\right)\cr &&\quad\quad- 6\left(10^{50}+2\times 10^{25}+1\right) \cr
&=& 10^{100}+4\times 10^{75}-8\times 10^{25}-5
\end{eqnarray} $$ It might seem 'obvious' that this value is greater than $10^{100}+1$ but it is a good idea to get into the habit of writing down precisely why it is obvious - the problem here is that we are suggesting the that sum of three different terms is greater than $1$. A simple way forward is to write out an inequality where we simplify one of the terms at each stage until we are left with just two terms as follows:
$$
\begin{eqnarray}
4\times 10^{75}-8\times 10^{25}-5&> & 10^{75} - 8\times 10^{25}-5\cr
&> & 10^{75}-10^{26}-5\cr
&> & 10^{75}-2\times 10^{26}\cr
&> & 10^{75}-10^{27}\cr &> &1 \end{eqnarray}
$$
(Note: you might not see the 'point' of these inequality manipulations: they are useful because it is clear and easy to verify each individual step. This turns something which might just be controversial into something that is not at all controversial.)

Now for writing out the number $N$ on the left hand side of the inequality. As the number is so large a computer or a spreadsheet will not easily help us. Keeping the '+1' part separate for as long as possible gives us (where $X=10^{25}$)

$$ \begin{eqnarray} N &=& (X+1)^4-6(X+1)^2 \cr &=& X^4+4X^3+6X^2+4X+1 - 6(X^2+2X+1)\cr &=& X^4+4X^3-8X-5 \end{eqnarray} $$ The structure of this number is now clear to us and includes the problematic negative terms, which we can deal with one at a time.

 

Firstly, $8\times 10^{25}$ is represented as $8$ followed by $25$ zeros. Removing this from the part $4\times 10^{75}$ leaves a number of the form

$$
100\dots 0039\dots 9200\dots 00
$$
Removing the $5$ from this gives a number of the form
$$
100\dots 0039\dots 9199\dots 995
$$
The trick is now to get the $1$s and the $3$ in the correct place. If we consider the number as a string of digits, counted from the right, then the first $1$ with be in the 101st place, the $3$ in the 76th place and the second 1 in the 26th place.

10,000,000,000,000,000,000,000,003,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,919,999,999,999,999,999,999,999,995

Phew! Who would have thought that place value could be so tricky? If you are planning on entering a career in finance or science then part of your computer programming will be to ensure that large numbers that you enter into your code are accurate. A lot could rest on this accuracy, so patient and careful detail are the key skills required.