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

1 1 1 1

Let +++ &c. ad infinitum =s.

1 3 6 10

That is, (1-1)+(1-1)+(1-1)+(1-1)&C=

11. And if it be

of the same serie

Let z=

Then z

; or s=2= sum required.

required to find the sum of ʼn terms .1 1 1 1 1 tatat 3 6 10

+

1 1 1 1

15

&c.

+=+=+=+= &c. to
4 5

1 1 1 1

+=+=+= &c. to

2 3 4

*|*|

2n

1 1 1 1 1

+=+=+⋅ + -&c. to

2

3 6 10 15 n(n+1 n+1

of the series, as was required.

1

sum of n terms

12. Required the sum of the infinite series,

2.3.4

1 1 1 1 1

Let z= +=+=+=+= &c. ad infinitum,

Then

1 2 3 4

1 1 1 1

2

+ ++ &c. by transposition.

1 2

Or

1

+ + + &c. 1.2.3 2.3.4 3.4.5' 4.5.6

And÷2= 2.3+3.4+3.4.5+ &c.

13. And if it were required to find the sum of n terms

1

1

1

1

of the same series +: +: + &c.

.2.3 2.3.4 3.4.5 4.5.6

1

Whence

1

1

1

&c. to

4 2.(n+1)(n+2) 1.2.3 2.3.4 3.4.5
+ 2.3.4 +
1.2.3+
in terms, by division.

1

-32

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

sum required.

14. Required the sum (s) of the series

And z=(1+x)x(x-x2+x3-x2+x5&c.)

Whence, by multiplication,

x-x2+x3-x4+x5 &c.

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

Therefore z=x, and x-x2 + x3-x4 + x5 &c. =.

1

1+x

sum required.

14
1+1 3

1 2 3 4

2 4 8 16

15. Required the sum of the series +++. +

z=

1

Let x=- and s=

2

-=x+2x2+3x3+4x4+5x3 &c.

And (1-x) × (x+2x2+3x2+4x+5x5 &c.)

Whence, by multiplication,

x+2x2+3x3 + 4x4 &c.
1-2x+x3

x+2x2+3x3+ 4x4 &c.

-2x2-4x3-6x3 &c.
+x3+2x4 &c.

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

1

4

9

16. It is required to find the sum (s) of the series+

Then

=

(1-x).

2

3 (1 = x) 3

=S.

-=x+4x2+9x3+16x+25x5 &c.

And 2 (1-x)3x(x+4x2+9x3+16x &c.)=x+x2, as will be found by actual multiplication.

17. Required the sum (s) of the series

a+da+2d, a+3d ·+. + +

mr

mra Mr3

&C...