Page 57 - Foundations Of Differential Calculus

40 2. On the Use of Differences in the Theory of Series
and

1 1 2 1
S. x = x + x,
2 4 4
so that

2
3x + x 1 3 2 1 1 2
S. = x + x + x = x (x +1) ,
2 2 2 2
5 2
and this is the desired partial sum. Thus if x =5,wehave 2 · 6 = 90,
while the sum of the terms is

2+7+15+26+40 = 90.

Example 2. Find the partial sum of the series 1, 27, 125, 343, 729, 1331,
... , which is the sum of the cubes of the odd integers.
The general term of this series is

3 3 2
(2x − 1) =8x − 12x +6x − 1,
so we collect the partial sums in the following way:

3 4 3 2
+8.S.x =2x +4x +2x ,
2 3 2
−12.S.x = −4x − 6x − 2x,
2
+6.S.x =+3x +3x,

and
0
−1.S.x = −x.

Then the desired sum is
4 2 2 2
2x − x = x 2x − 1 .

Hence, if x = 6,wehave36 · 71 = 2556, which is the sum of the first six
terms of the given series:

1 + 27 + 125 + 343 + 729 + 1331 = 2556.

65. If the general term is a product of linear factors as in paragraph 32,
then it is easier to find the partial sums by the method treated in that