m

.

m+27 +m+387&c. to n terms +

m

m)) +m

m

mim+y)+

mr

mr.

+&c. Here, leaving out the numerators, and the last factor of each denominator, let there be assumed 1 1 1

1

1 + ta

+
t

Then '

m+n 1)r mtnr 1 1 1

1

1 +

+ m+2r 3r

mt nr Whence, by subtraction and substitution, +

+ &c. to n terms m(m+ r) (m+r)(m+2r)" (m+2r)(m + 3r) 1 1

1
= sum required. Also
m+nr
1
1

1

= (m+r)(m +25) +(m+2r)(m+36) + &c. to n terms =

(3r 1

Where, if the numberTM of terms (n) be increased Mr + nepl"

1 without limit,

will vanish, and the sum of the series 1

mrt nr will be

Or, which is the same, = the reciprocal of the product of the two quantities m and r. In which case, if m and r be each taken

1 1 1

1 : 1, we shall have

1.2 +2.3 +34 +16 + &c. to n terms 1 1

1+ni+ni And by taking mand r of different values, the sum of in terms of a variety of other series may be found. 139. Required the sum of the infinite series 2r

2r
mim+r)(m+2r) +(m+r)(m+2r)(m+31) +
2r

+ &c. Here, leaving out the nume(m+2r)(m+- 3r)(m + 4r) rators, and the last factor of each denominator, as before, let there be assumed 1 1

1 +

Then mim+") *(mtr)(m+2r) +(m+2r)(m+3r) +-&c. = 8. 1 1

1 + (mtr) (m +2r) (m+2r) (m + 3r) (m+3r) (m+4r) 1

Whence, by subtraction, m(m + ) 2r 2r

1 mlm Ft)(m+27) + (mto)(m +2r)(m+31) +-&c.

mom tr) 41*

+1.5

n

+

1.2.3 +2.3.4 +3.2.67&c. to n terms, –

1

1 Also

mim+r)(m+2r) + (m+r)(m+2r)(m+3r) +&c. =
1

In which case, the sum of n terms of the series, 2rm (m + 2r) determined as in the 139th will be = 1

1

So that if n and i be 2rmim+r) 2r(m+nr){m+(n+1)=}

r"

1 1 1 each taken 1, we shall have

1.2.3 +2.3.4 +3.4.6+&c. ad in1 1 1

1 finitum =

And 4'

4 1

And by taking m and r of different values, 2(n+1)(x+2) the sums of various other series of this kind may be found, as in the preceding example. Questions without Solutions.

140. Sum 100 terms of the series 2, 5, 8, 11, 14, &c. and 50 terms of the series 1+2+3+44+5++&c.

Answers, 15050, and 42925. 141. Required the sum of the infinite series 1+3x+6x? + 10 +15x +&c. when x is less than 1.

142. Required the sum of the infinite series 1+4x*+10° +20%* +35x++&c. when x is less than 1.

1 1 1 1 1 143. Sum the infinite series

1.3

+ 9.11

+ 1 1

5 1 &c.

Ans.

; (1-2)' (1-2)* 10 or 2

8 ' 144. Sum 40 terms of the series (1/2) + (2X3) + (3X4) + (4x5) + &c.

Ans. 22960.

2x = 1 2x - 3 2x-5 145. Sum n terms of the series + + + +

22

2x 22 21- 7.

1
1

1
the infinite series
20

1.2.3.4 +3.4.5.67&c. and n terms of the series 1+8x+27x* +64x8+125x*, + &c.

1 2x

(1-x)' 146. Sum n term of the series 1 2 3 4 5 6

1

nr+
ga to two
25

(t-1)

])?? 147. Required the sum of n terms of the series 1 1 1 1

1 1.4

+ &c. +

4.7 m(37)

+3.5

+57 +7.9

+

1–23

; and

+2.3.4.5

n

i īg; and

; and 1+4n+2

18

Answers, n( + + + + + + &c. Ans.

1

+2.5 +3.6

+

n

n

n

tas

to.

+

n

)

+

+

S

n

n

An.
3+ 3n+1246n +277-9m

++9n

:

1 1 1 148. Required the sum of the series 4.8 +6.10

+&c...

2.6 1

3

5n+3n +

Ans. 85 2n(4 + 2n)

16

32+48n+16n? The symbol , in this and the following series, denotes the sum of an infinite number of terms, and s the sum of n terms.

1 1 1 1 149. Required sum of the series

3.8 " 6.12 9.16 12.20 1

1 +&c. . . +

Ans. E= 3n(4+4n)

12 12+ 12n

6 6 6 150. Required the sum of the series

t +

2.7 7.12 12.17 6 5

3n 17.227&c. . . +(50—3).(5n+2)*

Ans. 2

5

2 + 5n 1 1 1

1 151. Required sum of the series

十 4.8

6.10 8.12 10.14 1

1 + &c.I

Ans. (2+2n)(6+2n)

48

16+-16n 36+24n 1 1

1

1 152. Sum the series + +

+&c.+ 8.18 10.21 12.24 14.27 1

3

Ans. (6+2n).(1573n)

80' 24(8+2n) 730(10+2n) 1 1 1 1

1
153. Sum the series
3.6

&c. 6.8 + 9.10

12.12 15.14 1

1

Ans. = +3n(4+ 2n)

24

2(3 + 6n) 4(6+6n) 1

1 1 1 154. Sum n terms of the series

1.2.4 +2.3.5 +3.4.6 +4.6.7 1 &c. +

Ans. =(1+x) (3+7) 3(1+x) 12(2PN) 18(34)

n

2 3 4 5 155. Required sum of the series,

3.5+5.7-79-90+ &c. 1+n

1
1

1
Ans.
· +1+3n).(3+2n)

12' 124(3 + 4n) 2 1/ 3 1

5 166. Sum the series

3 1.3 3.5 27 5.7 1+n

1

1
Ans. E
3n(2n-1).(2n+1)

n

n

n

n

+ &c.tn(1+n).(? + n)

X = 51. ; Ans.a

= 7.

Z

5 6 7 8 156. Required, sum of the series

1.2.3 +2.3.4 +3.4.5 +4.5.6 4+n

3 3 2 1 Ans. S=ž

2 i+n+2+1

itn 2.1 + VC

2x157. Given

375

to find the values of 2.0 -NE 2x+Nx

Ans. =4. z by a simple equation.

75+1. 158. Find the cube root of 4N5+8.

Ans.

32 4x 34 258-52 69-X 159. Given

; to find the value of 1 17

3

2
4.x 2 2x +11 7 - 8x
160. Given 2x

13
5

7 161. Let abor+(1+c)bd Nc+cber}={bdNc+(ab+c)(1+c)}z. to find the values of x. By tr. 6°(ab+c)2 —{bdNc+(ab+c)(1+c)}x=-(1+ c)bd.Nc. Let x =

2-{BdNc+(ab+c)(1+c)}z=-bod(att 8°(ab + c)' c)(1+0)wc. Let m=bdx c, and n=(ab+c)(1+c), then will Nc

n 24-(m+n)zamn, by art. 70, case 4, I have =^{(m_2mn

z{ +n?)}#}(m+n); bab+c).c

, or z=m, or n; BNC bd NC

(ab+c)(1+0)

; and 2 = l'(ab + c) abtc

b*(ab+c) 72 162. A merchant having mixed a certain number of gallons of brandy and water, found, that if he had mixed 6 gallons more of each, there would have been 7 gallons of brandy to every 6 gal. lons of water ; but if he had mixed 6 gallons less of each, there would have been 6 gallons of brandy to every 5 gallons of water: how many of each did he mix ?

Let 70—6, and 6x - 6 denote the number of gallons of brandy and water respectively; which assumption fulfils the first condition of the question. By the second condition we have 72—12: 6x12::6: 5, or x : 62–12::1:5, whence 5x=62—12, or x=12; and 72—6=78 gallons of brandy, 62–6 = 66 gallons of water.

163. Find two numbers, such, that their sum, product, and difference of their squares, shall be all equal.

The sum of two numbers differing by unity, is equal to the difference of their squares. Let, then, 2+1 and x denote the num. bers : then their sum must be equal to their product ; that is, 2x+se+r, or r=by case 2; 4.c --4x+1=4+1=5 extracting the root, 2x_=15, .. +15; and ztl

m=i= +5, the two numbers required. 164. Find two numbers, such, that their product may be equal

orx

5 2

+

= 3.2

X = 3.

to the difference of their squares, and the sum of their squares equal to the difference of their cubes. If we assume ax and (a +1)x for the numbers, a being = 4+15, we shall fulfil the

= first condition of the question. (See Quest 163.) By the second condition, we must have (2a? +-2a+1)x==(30++3a+-1)xø ; 20'+2a+1 4+3

(since a = a +1).
?

a + 1). Hence, ax = 3a? +3a+16+4' 4a+3a 7a+ 4 ? 7a+ 4_4+IN5 V5(35+7)

15, 6a+4 6a+4

3,5+7 and (a + 1) = a*x = ax Xa=in 5(1 +115)={(5+15):

). that is, the numbers are 15, and 1(5+5). 165. Given 11 (6x+8) — 1(5x+3)=$(27 — 4x) — 2(3x +9), 166. Given x+

27—9x 5c+2_61 2x+5_29+4x
+

to find
4
6 12 3

12 the value of x.

Answers, x = 6, and x = 5, 72-8 15x+8

31167. Given

to find x.
11
13

2
168. Given }(5x— 1)- 40172-2)=61-2, to find 2.
ба+10aь?

5N (a+b), (1+ 262)cdN od 169. Given

373
+
3 a

ab Xn(a+b)c, to find the values of x.

a
,

262)cdNC By tr.

. 36(3-2)

363

3 a
1+202 5N (a+b)
; and na

then will
ab
3 –

303 5am

cd NCN (ato)

3622 -2_{n+mcdx c}x=362

бат 5cdmx CN(a+b)

5am 3_(n+mcdx c) =

; .. 363

= Z, or ?

36 (n + mcdNC)z =- cdmnnc. By case 4, art. 70, I shall have {z-i(n+mcdx c)}={(n+mcdc)' – cdmnx c; or {2}(n+incdNc)}=tn_mncdə c+4(moc°dc); root =2 i(n+mcdw c)—+1(n-mcdwc). Hence I have

5a(1+262) z=f(n+mcdc)}(n—mcdc);

= n, or 36 (3 — a?) 5N (a+b) (1+262)cd NC mcdNC, ..z=

;

; hence
363

3-a?
(3—?)(a+b).
anb 36'cd nc ,

Answer.
ab(1+262)

5a 170. Find the cube root of 26+1573.

Ans. 2+ 3.

dag) 5a(1 +26°), 58 (+0) (17 2390 ) cdxCN(a+b). Let ma

; let 2

+

a

; then

ab

X, or Z

or

; or