+ &c.

9

Giot

)

Or, 10+

10000+&c.

S

10007&c.

125. Required the sum (s) of the circulating decimal:999999... continued ad infinitum.

9 9 9 9 Here .999999

10
+
100
+1000+

10000
1

1 1 1
10 100 +1000 it + &c.

10000
1 1 1

1 1 1000+ 10000

::1+10+

+ 100

9

100 1 10s

10s

9s and ... 15

ES ; whence s== 9

9 9 sum of the series.

126. Required the sum (s) of the series ał(a+d)?+la+20)+ (a+30)2+(a +4d)?+ &c. continued to n terms.

Here a'=a; (a+d)=a?+2x lad+ld? (a+20)=a+2x2ad44d* ;' (+34)=a2+2X3ad+9&?; and (a+40)+2x4ad716d ; whence

Sum of n terms of (1+1+1+1+ &c.)a?

ditto of 0+12+3+ &c.)2ad

ditto of (0+1+4+9+ &c.)d? But n terms of 1+1+1+1+ &c. = n. And of 0+1+2+3+4

m(n-1) +&c.

Also of 0+1+4+ 9 + 16 + 25 + &c 1.2 n(n-1)(2n-1)

; .. sana+n(n-1)ad+ 1.2.3

1.2.3 the whole sum of the series to n terms.

127. Required the sum (s) of the series aita+d)+(2+2d) +la+3d)7(a+4d)*+ &c. continued to n terms. Here a =

(a+ d)=a+3X la’d+3X ladat 1
(a+20) =a + 3x2a'd3X 4ad 8d
(a+30)=a’+3X3ad73X 9ad++ 2743
(a+4d3a 13X 4a ́d +3X16ad" 6443
(a+5d)=a+3X5a%d73X25ad?+125d3
Sum of n terms of (1+1+1+1+&c.) a

.ditto of (0+1+2+ 3+&c.)3a'd Whence s =

.ditto of (0+1+4+ 9+&c.)3a’d

.ditto of (0+1+8+27+&c.) do
But n terms of 1+1+1+1- &c. = n.
Ditto ..... of 0+1+2+3+ &c.=\n(n-1)

+

n(n-1)(2n-1) Ditto ..... of 0+1+4+9+ &c.=

.

2.3

(-1)^2(x−1)*6(7–1) Ditto .... of 0+1+8+27+ &c. =

4

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

ar-a,

a

a

a

1xá

a

a

a

a

x{

n(n-1)3a'd, nin-1)(2n-1)3ad Therefore s=na+

( +

+ 2

6 (n-1)+2(x−1)3+(n-1)*

_d = sum required.

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=l and r=2, and we shall have atartar tard tar4+ &c.=1+2+4+8+ 16 + &c. But the successive sums of 1, 2, 3, 4, &c. terms of this series are

ar2 1. (n-1)x;

2.

(7.1)X ar2

art. 3. (78—1)X

(79—1)x

_1= 1-1

11 &c.

&c. Therefore sa mix

In terms of rtpol tpos tapet &c.

-n terms of 1+1+1+1+ &c. But atrofp3+r+&c.=(r"–1)X 1 and 1+1+1+1+1 =

+

r(got — 1) +1+it &c.=n. Whence s x —п Х

-1 2(2"-1)-N=whole sum required.

1 3 7 15 31 129. Sum n terms of the series i+ätáto

tota&c. 1

8 16 Here, the terms of the series are the successive sums of the ge

1 1 1 1 1 metrical series itatatäti + &c. Let, therefore a = 1

8 16

1 1 1 f and r=2; then will it ++8+ &c. =a+ =

. rattt

+++ 2 4

24 The successive sums of 1,2, 3, 4, &c. terms of this series are, (r--1) Xa

(78 — 1) Xa 1 1. =(-1) X 3.

(r(r-1)X1

(1 — 1X m2

I (p1) Xa

(741) Xa 1 2.

X

4. (1-1) X1

(r

1) Xp &c. &c. Therefore n terms of r+r+r+r+ &c.

.
1

1 1
|-n terms of it-titt &c.

But rtrtrtrtrtrt &c. = nt.

a

xa

4

4

a

200 X

a

n

S

S.

+

S

+3)+()+(

4)+ )+&c. =

4

S

=

1

.. to-.

1 1 1

go" - 1 And it, tet+ &c. =

Whence 2 pou3

(r — 1)p-°
(n-1)2+1
21-1

= sum required. (1-1) 130. Required the sum of the infinite series of the reciprocals of the triangular numbers ++itibtist&c. Let Hitt tistas &c. ad infinitum

1 1 1 1 1 Or

tistas tas +3.5+ &c. .... = s.

+
1.1 +2.3 +2.5

1 1 1 1 1
Then

That 1.2 +23+3.4 +45 +56 + &c....

2 1

1 is 2

តួ 2 3

2 1 1 1 1 1 1 1 t3 ta to

+5 to + + &c. Or, 1

- Whence 1 1 1

1

2

+ &c. 2 3 4 5 6 7 =t; or s = 2 sum required. 131. Find the sum of n terms of the same series, i + $

于。 1

1 1 1

1 +16+15+ &c. Let sitate to tát&c. tit &s=

++ 10

5 1 1 1 1 1

1

1
12 titats+&c. to .. to
+ +

+ 4 5 1 1 1 1

1 nt12 titatót&c. to Therefore, by subtracting 3 4

1 1 1,1 1 1 this from the first, we have

1 n+12

to. +ia +20+ &c.

十十十十 c
1 1
1 1 1 1

1
Or
nt1 n+1=2 te

tátio too + &c. to
6 12 20

7(71) 2n 1 i 1 1

2

1 1 Whence tota tint&c. to

Orität n+11

10
(m+1)

3 1 1 1

2

2n 6+

= sum of n terms of the

3(m+1) +1 series, as was required.

1 1 1 1 132. Of the infinite series

+ &c.

4.5.6 required the sum.

iti+*+*+++&c. ad infinitum. Then s-1=1+1+1+1+&c. by transposition.

1 1 1 And 13

+ &c. by subtraction. 4.5

2

4

n

Then s

to

+

c

+5.6.7

Let sa

+&c. to min+1)

1 1 1 1 1 Or 1

+

+ &c. by transposition. 2 2.3 3.4 4.5 5.6

1 4 6 8 Whence

2

1.4.3 +2.9.4 +3.16.5+ &c. by subtraction.
1 2 2 2 2
Or
2

1.2.3 +2.3.4 +3.4.5 +4.5.6 + &c.
1
1 1

1
And
2
1.2.3 +2.3.4 +3.4.6 by dividing each side by 2.

'
1 1
1

1 1 But

2

i .. + 2.3.4 +3.4.5 +4.6.6 &c. 2 4 1.2.3 ad infinitum , which is the sum required. 133. And if it were required to find the sum of n terms of the

1 1 1 1 same series +2.3.4 +3.4.5 +4.5.6

+

&c.
1.2.3
1 1 1

1
+ +
1.2 2.3 3.4
1 1 1 1

1
Then s

+ + + &c. to
2 2.3 3.4 4.5

7(z + 1)
1
1

1 1 1 1 1 1 And s

+
2 (m+1)(n+2)

(
2.3 +3.4

+
+4.5 +5.6 +6.7

7.8 1 &c. continued to

terms. Therefore (z

++?) 1 1

2 2 2 2 (n+1)(n+2) 1.2.3 2.3.4 3.4.5

+3.4.5 + &c. to n terms, by 1 1

1 1 subtraction. Whence

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

1.2.3 T 2.3.4 1

1 + &c. to n terms, by division. And, consequently, 3.4.5

1.2.3 1 1

1

1 + &c. continued to n terms = 2.3.4

4 2(m+1)(n+2) =sum required.

134. Required the sum (s) of the series 1-47f-Tot &c. continued ad infinitum.

Let X = 1 ; in which case sam **_** + &c. Then, if this be multiplied by 1+x, we shall have s(1+x)=(1+2) X (-??x4 +2&c.), where, by performing the operation on the right hand member, there will arise s(1+x)=x. Therefore, s=2—24+203_24+25_&c.

1 1 1

1/ 1
Or
+

-&c. = Fitx 2

16

32 required.

5.6 +6.77

+

+3.4.5

+

sum

1+13

十十十十十

sum.

SEX

=1}

9

135. Required the sum (s) of the series 3 + + + +

+ &c. continued ad infinitum. Let x =* ; in which case S=x+ 2x + 3x + 4.x* + &c.

Then, if this be multiplied by (1-2)', we shall have s(1-x)=(1-2)*X (2+2x++ 3x2 + 4x* + &c.) Where, by performing the operation at length, on the right hand member, as be. fore, there will arise s(1-2)=. Therefore, also, x+222 + 32% +4+* +506 + &c.

1 2 3 4 5 Or

去 (11) 2

+ 2

+is+32

toot&c.
8
4

(141) 136. Required the sum (s) of the series $ +ft+f+

+ &c. continued ad infinitum. Let x = $; in which case S= 2.+4r +92 +16* + &c. Then, if this be multiplied by(1-x), we shall have s(1 – t)

2 =(1–2):X (x+4x2+92 +16x +&c.)=+*, as will be found by performing the operation. Therefore s(1-2)=x+x, or se

xtxe x(1+x)

(1-2)*(1-)? 1 4 9 16

f(175) 3 Org+9+27+81+&c.

(1-3)

= sum required.

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

atd, a+2d, a+3d, a+4d +

+ &c. mp mpe

mrt 1

atd Let x=-; in which case s = + x + + &c. Then, if this be multiplied by (1–2)?, we shall have s(1 — 2)

a+2d tota

1+3d =(

-23 + &c.). Or, by

ad actually performing the operation, s(1-x)?? -*; and, by

a-a-d) by division, s=

m(1-x)? Whence, restoring the value of x, we shall have, by substitu

1

(-d) tion and reduction, s

ar? — (a —d)r

And, con1

mr-1)

m(1--) sequently, - +

+&c.=
"ar-1+d

(1-1)
sum of the series, as required.
138. Required the sum of n terms of the series

+

a

+

+

+

m

mr

a

a+2d

+

m

mim trs tim

tr)(m+2r) +(m+2r)(m+37) +(m+30)(m +47) (