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)

Climbing Powers

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?

Tens

Age 16 to 18 Challenge Level:
We received lots of great solutions for this problem, from Niharika at Rugby School, Lee from Garden International School, Joseph from Bingley Grammar, Nathan from Mossbourne Community Academy, James from Samuel Whitbread Academy, and Seemanta from St Placid's High School.

Most of you noticed that all the sums are multiples of ten; here is a solution sent in by Amir from Wallington Country Grammar School:

In order for a number to be a multiple of 10, it must end in a zero. We need to find out if $3^n + 7^n$ will end in a 0 in order to answer this question.

To do this we can construct a table to see what the 'units' digit is when we put either $3^n$ or $7^n$.

 n $3^n$ unit $7^n$ unit Sum end in 0? 1 3 7 Y 2 9 9 N 3 7 3 Y 4 1 1 N 5 3 7 Y 6 9 9 N 7 7 3 Y

It is evident that only the odd values of N give a multiple of 10, however we can see a pattern; the unit of the number will cycle after you add 4. For example, $3^1$ ends in the same digit as $3^5$ and $3^9$. As a result the final digit of the sum $3^n + 7^n$ will always be 1 of 4 possibilities. It is either '3 + 7', '9 + 9', '7 + 3' or '1 + 1', of which two are multiples of 10 and happen to be when n is either 1 or 3. Therefore we can see that adding any multiple of 4 to 1 or 3 will always yield an odd number - hence odd number values of n will always give multiples of 10.

We can extend this by saying that when n is in the form '4k + 2', the final digit will be an 8 (9+9=18) and when n is in the form '4k' (i.e. is a multiple of 4), it will end in 2 - either way it will be a multiple of 2. Therefore the sum is always even.

Note: k is any integer greater or equal to 0.

Daniel from Chetham's School of Music provided proofs of the same result using induction and the Binomial theorem. Here is his solution: .pdf

Finally, Michael noticed that this result can be generalised in another way:

By factorising generalised expressions we can prove all four of the given expressions are divisible by ten. For example,

$$x^{2m+1}+y^{2m+1}=(x+y)(x^{2m}-x^{2m-1}y+x^{2m-2}y^2-\dots -xy^{2m-1}+y^{2m})$$

Meaning that for any two numbers that add up to 10 (or any multiple of 10), $x^n+y^n$ is divisible by 10 for odd n. Thus this proves the first two given expressions divide by ten. Secondly, observe that

$$x^{2m}-y^{2m}=(x+y)(x^{2m-1}-x^{2m-2}y+\dots -y^{2m-1})$$

Meaning that for any two numbers that add up to some multiple of 10, $x^n-y^n$ is divisible by 10 for even n. This factorisation therefore proves the last two expressions divide by ten.

Well done everyone!