125. Required the sum (s) of the circulating decimal 999999...

continued ad infinitum.

Or, 10

100

+1000+

10000

+&c.:

+

126. Required the sum (s) of the series a2+(a+d)2+(a+2d)2+ (a+3d)2+(a +4d)2+ &c. continued to n terms. Here aa; (a+d)2=a2+2× lad+ld2

(a+2d)2=a2+2×2ad+4d; (a+3d)2=a2+2×3ad+9ď; and (a+4d)2=a2+2×4ad+16ď2; whence

8=

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

キ

. . . ditto of (0+1+2+3+ &c.)2ad
... ditto of (0+1+1+9+ &c.)ď2

But n terms of 1+1+1+1+&c. = n. And of 0+1+2+3+4

n(n-1)

1.2 n(n-1)(2n-1) 1.2.3

Also of 0+1+4+ 9 + 16 + 25+ &c

'; ... s=na2+n(n-1)ad+

the whole sum of the series to n terms.

127. Required the sum (s) of the series a3+(a+d)3+(a+2d)3 +(a+3d)+(a+4d)+ &c. continued to n terms. Here a3 a (a+ d)3=a3+3×1a2d+3× laď2+ 1ď (a+2d)=a3x2a'd+3X 4ad2 + 8ď3 (a+3d)3=a3+3×3a2d+3× 9ad2+ 27ď3

Whence s

(a+4d)3 a3+3x4a2d+3X16ad264d3
(a+5d)3=a3+3×5a2d+3×25aď2+125ď3
Sum of n terms of (1+1+1+ 1+&c.) a3
.ditto of (0+1+2+ 3+&c.)3a'd
..ditto of (0+1+4+ 9+&c.)3a'd
.....ditto of (0+1+8+27+&c.) ď3

But n terms of 1+1+1+1— &c. = n.
Ditto ... of 0+1+2+3+ &c.={n(n−1)

Ditto

.....

n(n-1)(2n-1)

..... of 0+1+4+9+ &c.= 2.3

Ditto.... of 0+1+8+27+ &c. =

(n—1)1+2(n—1)3+(n—1)3 4

128. Required the sum (s) of n terms of the series 1+3+7+ 15-31 &c. Here the terms of the 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 atar+ar2+ar3+ar1+ &c.=1+2+4+8+ 16 + &c. But the successive sums of 1, 2, 3, 4, &c. terms of this series are

a

2.

r-1

a

a

a

-1.

But a+rofr3+r*+&c.=(r”—1) ×———, and 1+1+1+1+1

X

-1'

3 17 15 31

129. Sum n terms of the series+++ + -&c.

8 16

Here, the terms of the series are the successive sums of the ge

The successive sums of 1, 2, 3, 4, &c. terms of this series are,

(n-1)2"+1

=

27-1

=sum required.

130. Required the sum of the infinite series of the reciprocals. of the triangular numbers +++i+is+&c. Let +is+is+&c. ad infinitum = s.

1

1

1

1

Or +1.3 +23 +2.5 +3.5 + &c. .... = s.

1.1

1

1

Then

+ &c..

....

+

5

+&c.

3

1

3

1

+ + + + &c.

5 6

1 1 1

; or s2 sum required.

131. Find the sum of ʼn terms of the same series, + § ÷ 1

+ +15+ &c. Let s=1+2 to t2 to t&c. to-.

1 1 1 1

4 5

2

4 5

1

10

1

1

Then s

+++&c. to . . to

And s

+

n

Therefore, by subtracting

1 1 1

6. 12 20

1

1

1

1 n

toot &c.

2

Then s = 3 + 3 + 4 ++&c. by transposition.

1

And 1: 1.2+2.3 +3.4 +45 + &c. by subtraction.

sum of n terms of the

1

1

1

+&c.

2

1.2.3 +2.3.4 +3.4.5 +4.5.6 + &c.

But÷2=4

1

1

1

1.2.3 +2.3.4 +3.4.5 by dividing each side by 2.

1

1

1

123 +2.3.4 +3.4.5 +4.5.6 +5.6.7

4, which is the sum required.

&c.

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

1

1

1

terms.

(n+1)(x+2)

4.5 +5.6 +6.7 +

Therefore

1

7.8

2 ̄(n+1)(n+2)=1.2.3 2.3.4 +3.4.5+ &c. to n terms, by

+ &c. to n terms, by division. And, consequently, 1.2.3

1

+ + &c. continued to ʼn terms =

2.3.4 3.4.5

sum required.

1

1

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

134. Required the sum (s) of the series-+-+6+ &c. continued ad infinitum. Let x= ; in which case s=x-x x3-x2+ &c. Then, if this be multiplied by 1+x, we shall have s(1+x)=(1+x)× (x—x2+x3-x2+x-&c.), where, by performing the operation on the right hand member, there will arise s(1+x)=x. Therefore, sx-x2+x3-x2+x3-&c.

135. Required the sum (s) of the series ++&++ +&c. continued ad infinitum. Let x; in which case s=x+2x2 + 3x3 + 4x2 + &c.

Then, if this be multiplied by (1-x), we shall have s(1—x)2=—(1—x)2×(x+2x2+ 3x2+4x2 + &c.) Where, by performing the operation at length, on the right hand member, as before, there will arise s(1-x)2=x Therefore, also, x+2x2+3x3 +4x2+5x+ &c.

x

(1-x)2'

Or

1 2

3 4 5

2 4 8 16 + 32+ &c.

136. Required the sum (s) of the series

=

sum.

(1—1) 2 ++++ 25 &c. continued ad infinitum. ́ Let x = ; in which case s=x+4x2+92 +16x+ &c.

Then, if this be multiplied by (1-x)3, we shall have s(1 — - x)3 —(1—x)3× (x+4x2+9x2+16x*+&c.)=x+x2, as will be found by performing the operation.

Therefore s(1-x)=x+x2, or s=

(1—x)3 (1 —x)2°

3 9 27 81

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

+a+da+2da+3d a+4d

Let x; in which case s= + -x+ -x2 + &c.

Then, if this be multiplied by (1-x)2, we shall have s(1 — x)2.

Whence, restoring the value of x, we shall have, by substitu

sum of the series, as required.

138. Required the sum of n terms of the series

(r−1)2

m(m+r) +(m+r)(m+2r) +(m+2r)(m+3r) +(m+3r)(m+4r)

m

{a(r=-1)

a(r-1)+d