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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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Squaresearch

Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

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Loopy

Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?

Double Trouble

Stage: 4 Challenge Level: Challenge Level:1

Charlie has been adding fractions in the sequence $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \dots$ where each fraction is half the previous one:

$$\frac{1}{2} + \frac{1}{4} $$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8}$$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} +\frac{1}{16}$$

 

Work out the answers to Charlie's sums. What do you notice?

Will the pattern continue?
How do you know?

Try writing an expression for $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n}$$

Could you convince someone else that your expression is correct for all values of $n$?

Charlie drew a diagram to try to explain what was going on:
 

 

Use Charlie's diagram to explain why $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^n} = 1-\frac{1}{2^n} = \frac{2^n-1}{2^n}$$

 

Alison has been adding numbers in the sequence $1, 2, 4, 8, \dots$ where each number is twice the previous one:

$$1 + 2$$ $$1 + 2 + 4$$ $$1 + 2 + 4 + 8$$

Work out the answers to Alison's sums. What do you notice?

Will the pattern continue?

How do you know?

Try writing an expression for $$1 + 2 + 4 + \dots + 2^n$$

Could you convince someone else that your expression is correct for all values of $n$?

Alison drew a diagram to try to explain what was going on:
 



Can you use Alison's diagram to explain why $$1 + 2 + 4 + \dots + 2^n = 2^{n+1}-1$$