Whence s = 1= sum of the series.

6. Required the sum (s) of the series a2 + (a + d)3 + (a + 2d)2 + (a + 3d)3 + (a + 4d)3, &c., continued to n terms. Here,

8 =

First, a2

= a3
(a + d)2 =
(a + 2d)2 =
(a + 3d)2 =
(a + 4d)2

&c.

=

a2 + 2 x lad + 1d

a2 + 2 × 2ad + 4d3

a2 + 2 × 3ad +9ď2

a2+2× 4ad +16ď3,

&c.

Whence

Sum of n terms of (1 + 1 + 1+1+, &c.) a2,

+ .. ditto of (0+1+2+3+ 4+, &c.) 2ad3, +... ditto of (0 +1 + 4+ 9 + 16 +, &c.) d. But n terms of 1 +1 +1 +1 +, &c., = n.

And of 0+ 1+2+3+4, &c.,

==

n(n-1)

1.2

7. Required the sum (s) of the series a3 + (a + d)3 + (@ +2d)2 + (a + 3d)3 + (a + 4d)3, &c., continued to n terms. First, a3 a',

S=

(a + d)3 = a3 + 3 x la'd + 3x lad2 + 1d3,

Sum of n terms of (1+1+1+1, &c.) a3,
+... ditto of (0 +1 + 2 + 3 + 4, &c.) 3a2d,
+ ditto of (0+1+4+9+16, &c.) 3ad,
+ .. ditto of (0 + 1 + 8 + 27 + 64, &c.) ď3,
But n terms of 1 + 1 + 1 + 1 + 1, &c., =n.

Ditto... of 0+1+2+3+4, &c,.

ノ

n (n - 1)

Ditto... of 0+1+4+9+ 16, &c.,

1

Ditto... of 0+ 1+8 +27 + 64, &c.,

8. Required the sum (s) of n terms of the series 1+3+7 +15+31, &c.

The terms of this series are evidently equal to 1, (1 + 2), (1 + 2 + 4), (1 + 2 + 4 + 8), &c., or to the successive sums of the geometrical series 1, 2, 4, 8, 16, &c.

Let, therefore, a = 1 and r = 2, and we shall have

a + ar+ar2 + ar3 + ar1, &c.,

1+2+4+8 + 16, &c. But the successive sums of 1, 2, 3, 4, &c., terms of the

a

r- 1

But 1+

Therefore, s=

n terms of r+ p2 + p3 + p1, &c. -n terms of 1 + 1 + 1 + 1, &c.

+1+1+1 +1 + 1, &c., = n.

whole sum required.

9. It is required to find the sum of n terms of the series

3 7 15 31 63

+-+ + + &c.

2 4 8 16 32'

Here the terms of this series are the successive sums of the

But the successive sums of 1, 2, 3, 4, &c., terms of the

These being the two series derived from the above ex

pressions;

But r+r+r + r + r + r, &c., = nr.

required.

+ + &c., =

(r — 1) pn

=

10. Required the sum (s) of the infinite series of the recipro

cals of the triangular numbers

1 1 1 1 Let + + +

1

3 6 10'

1

1 1 1 1

1

+ + + + &c. 1 3 6 10 15'

&c., ad infinitum = ́s.

1

Or, + + + &c. ...
1.1 1.3 2.3 2.5'

=S.

11. And if it be required to find the sum of n terms of the

1 1

+ + + + &c.

1

10 15'

1

ad infinitum, =

which is the sum required.

4'

13. And if it

the same series

were required to find the sum of n terms of

+ + +
2.3.4 3.4.5 4.5.6'

+ + +, &c., to

=

2.3 3.4

1

2 2.3 3.4

1

(n + 1)(n+2)

1

&c.

1

n (n+1)

1

+ +

&c., to

n (n + 1)°

+

2.3
1

+ +
3.4

(n + 1)(n+2)

2 (n + 1)(n+ n)

to n terms, by subtraction.

2

+ 4.5 5.6

+ +

1.2.3 2.3.4 3.4.5'

terms.

2

2

&c.,

4 2 (n+1)(n+2)

&c., to n terms, by division.

1.2.3 2.3.4
1

4 2 (n + 1)(n+2)

14. Required the sum (s) of the series

x3 + x1, &c.)

1 + x

And = (1+x) × (x − x2+x3 — x2 + x3, &c.)

Whose sum is = x+0+0+0+0, &c.