This page has been labelled as Series C4 and FP1, which tells you, perhaps, the expected standard. This page considers the first few instances of Σrⁿ.
The sum of the counting numbers can be viewed as an AP with a=d=1, so un=1+n-1=n and
Sn=n/2 {2+n-1} = n(n+1)/2. Alternatively, pair off the first and last (hence n=1) and count the pairs, n/2, => Sn= n(n+1)/2. This is often written as Σr = ½ n(n+1). The series itself is {1, 3, 6, 10, 15, ...} also called the triangle numbers.
Inductive proof 1:
Observe that for n=1, the formula gives 1 and for n=2 the formula gives 3. These are correct.
Suppose the formula works for n=k, Sk= ½ k(k+1). then
Sk+1= ½ k(k+1) + k+1 = ½ [k(k+1) + 2k+2] = ½ (k+1)(k+2) which is the structure of the formula for n=k+1, so the formula works for all k and therefore all n.
The sum of the squares of counting numbers, Σr² is a series {1,5,14,30,55,91,140,204....}. Observe the factors of these:- {1, 5, 2x7, 5x6, 5x11, 7x13, 10x14, 12x17...} Assume that 2n+1 is a factor {3, 5, 7, 9, 11, 13, 15, 17} and multiply the whole series by 3 to make it start right, i.e. 3 Σr² = {1x3, 3x5, 6x7, 10x9, 15x11, 21x13, 28x15, 36x17...} and then see that the initial term in each product (now) is a triangle number or Σr, whose formula we already know as ½ n(n+1), so our complete formula must be that Σr²=⅙ n(n+1)(2n+1).
Inductive proof 2:
Observe that for n=1, the formula gives 1x2x3/6=1 and for n=2 the formula gives 2x3x5/6=5. These are correct.
Suppose the formula works for n=k, Sk= ½ k(k+1)(2k+1). then
6Sk+1= k(k+1)(2k+1) + 6(k+1)²= (k+1)[k(2k+1) + 6k+6] = (k+1)(2k²+7k+6) = (k+1)(k+2)(2k+3) which is the structure of the formula for n=k+1, so the formula works for all k and therefore all n.
So to Σr³ : the series is {1, 1+8, 9+27,36+64, 100+125, 225+216,...} which adds up to become {1, 9, 36, 100, 225, 441, ...} which is observed to be {1, 32, 62, 102, 152,...} which are the squares of the triangle numbers, so the formula is Σr³ = ¼ n²(n+1)²
Σr⁴ is more difficult : the series is {1, 1+16, 17+81, 98+256, 354+625, 979+1296,...} which adds up to become {1, 17, 98, 354, 979, 2275, ...}. Looking in this numbers for prime factors we can see {1, 17, 2x72, 2x3x59, 11x89, 52x7x13,... Observing that previous series solutions have been multiplied by 2,4,6 we might try inspecting 8 Σr⁴ but that soon proves unhelpful, so it is better to revert to assuming first that the solution formula has factors of n and (n+1); the fifth and sixth terms have a 2n+1 factor, so assume the formula is of the form
P Σr⁴ = n(n+1)(2n+1) f(n) which forces the situation (and, using the dot for a multiply sign)
30 Σr⁴ = {1.2.3.5, 2.3.5.17, 3.4.7.35, 4.5.9.59, 5.6.11.89, 6.7.13.125, ...} and defines f(n) as the series {5, 17, 35, 59, 89,125, ...}. Finding a formula for this requires a different sort of theory, such as described in the Lower School page Formulae; the second difference is 6 and the zero term is -1, so the formula for f(n) is soon seen to be f(n) = 3n²+3n-1.
Exercise: predict Σr⁵, Σr⁶, Σr⁷, Σr⁸, Σr⁹, Σr10 and then check your predictions.
More to come if someone asks...
DJS 20130418
I confess here to looking up ‘series ‘ in a maths dictionary yesterday 20151209 and only then discovering that, by definition in maths, a series is the sum of a sequence. I remember discussing the confusion and interchangeability of the words sequence and series with many classes and wish I had been more clear, or more definite or more clearly definite. Apologies to all.
This comment refers to Σrⁿ: I expect the odd n to be simple and the even n to be difficult. I made a spreadsheet to keep the arithmetic from being a source of problems. Developing a general rule for the coefficients leads you to new pastures where I suspect you might find the footprints of Gauss. DJS 201600602
