Page 56 - Foundations Of Differential Calculus

2. On the Use of Differences in the Theory of Series 39
63. In these formulas we can observe a general law, with whose help we
can easily find any one of the formulas from the preceding formula, except
for the last term, in case the power of that term is 1. In that case we have
to find one more term. If we ignore that term for the moment, and if

n
n
S.x = αx n+1 + βx + γx n−1 − δx n−3 + x n−5 − ζx n−7 + ηx n−9 − ··· ,
then the subsequent formula will be
x+1 n +1 n+2 n +1 n+1 n +1 n n +1 n−2
S.x = αx − βx + γx − δx
n +2 n +1 n n − 2
n +1 n−4 n +1 n−6 n +1 n−8
+ x − ζx + ηx − ··· ,
n − 4 n − 6 n − 8
and if n is even, we obtain the true formula. If n is odd, then the formula
is lacking a term, which has the form ±φx. Now with very little work we
can discover φ. Since when we let x = 1 the partial sum should be but a
single term, the first term, and this is equal to 1, if we let x = 1 in all of the
terms found so far, the sum should be equal to 1. In this way we evaluate
φ. Once this is found we can proceed to the next step. In this way all of
these sums can be found. Thus, since

5 1 6 1 5 5 4 1 2
S.x = x + x + x − x ,
6 2 12 12
we have

6
3
5
· x +
S.x = 6 1 7 6 1 6 6 · 5 x − 6 · 1 x + φx,
· x +
7 6 6 2 5 12 3 12
or
6 1 7 1 6 1 5 1 3
S.x = x + x + x − x + φx.
7 2 2 6
1 1 1 1 1 1 1
Now let x = 1, so that 1 = 7 + 2 + 2 − 6 + φ, and so φ = 6 − 7 = 42 , just
as we found in the general form.
64. By means of these formulas for partial sums we can easily find the
partial sums of all series whose general term is a polynomial, and much
more expeditiously than by the previous method using differences.
Example 1. Find the partial sum of the series 2, 7, 15, 26, 40, 57, 77, 100,
2
126,... , whose general term is (3x + x)/2.
Since the general term consists of two members, we find the partial sum
for each of them from the above formulas
3 2 1 3 3 2 1
S. x = x + x + x
2 2 4 4