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54 2. Evaluation, Division, and Expansion

Taking notice of the fact that the nth term of any series is equal to

the difference between the sums of the first n terms and of the first n - 1

terms, one can see that the two formulae given above can be established

by verifying the algebraic results

n = +(n + 1) - i(n - 1)n

n2 = &(n + 1)(2n + 1) - i(n - l)n(2n - 1).

With these formulae which express the general term of the series as a

difference, we can obtain the required sum by “summation by differences.”

For example,

1+2+. ..+n= il.2, i2.3 -3.2 +...

>

+ in(n+l)-i(n-1)n

1

= $(n+l)

after a cancellation of terms. This method seems unsatisfactory since we

do not usually know in advance how to find a function whose differences

are n and n2.

To get around this, we try to find functions which have differences which

are simply described, and then try to express the summands of the series

in terms of these functions.

Let g(n) be a function of the integer n. The first order difference of g(n)

is defined by

b(n) = s(n + 1) - s(n).

Find Ag(n) when g(n) = nk (Jc = O,l, 2,3,4,5)? Verify that An6 = 6n5 +

15n4 + 20n3 + 15n2 + 6n + 1. The result is quite complicated. A function

which has a difference of a simpler type is the factorial power of n. This is

defined as follows:

let k be a positive integer. Then the k2h factorial power of n is

given by

rack) = n(n - l)(n - 2)... (n - k + l),

where there are k factors, each 1 less than its predecessor.

Verify that An(“) = kdk-‘).
The value of this formula is that we can now conveniently sum by differ-
ences. Using the fact that

(r + l)(“+l) - p(k+l) = (k + l)rck),