### Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

### Growing

Which is larger: (a) 1.000001^{1000000} or 2? (b) 100^{300} or 300! (i.e.factorial 300)

### How Many Solutions?

Find all the solutions to the this equation.

# Climbing Powers

##### Stage: 5 Challenge Level:

Why do this problem?
The problem demonstrates that mathematical language needs to be very precise and it is important to avoid ambiguity.

If the learners are encouraged to do as much as they can without looking at the hint most of them will discover for themselves that $2^{3^4}$ is ambiguous because $2^{(3^4)}$ is not the same as $(2^3)^4$.

The problem gives opportunities for all the learners to have some success and for them to investigate the powers of $\sqrt 2$ using a calculator or computer and discover the behaviour of the two different sequences for themselves.

At the same time the problem provides a challenge for the most able students to prove the results which calls for an understanding of functions. One proof uses the logarithm function and the other mathematical induction.

Possible Approach
After the class has worked on indices this problem gives reinforcement material for everyone and some extension possibilities for the most able.

The hint gives the learners a good deal of 'scaffolding' thus enabling the teacher to set the problem for the learners to tackle independently without a lot of introduction and help from the teacher, at least initially.

The problem could be set as homework with the intention of going over the proofs in class the next lesson and having a whole class discussion.

Key Questions
What does $a^{b^c}$ mean?
What is the difference between $(a^a)^a$ and $a^{(a^a)}$?
If you repeat (or iterate) the function which maps $a^a$ to $(a^a)^a$what is the next value? What is this function?
If you repeat (or iterate) the function which maps $a^a$ to $a^{(a^a)}$ what is the next value? What is this function?

Possible support

Try the probem Lastly...Well