.or 1 year, as before ; then that root of it which is denoted by the aliquot art, will be the amount of il. This amount being multiplied by the principal sum, will produce the amount of the given surn as required.

2d. When the time is not an aliquot part of a year. Reduce the time into days, and take the 365th root of the amount of 11 for year, which will give the amount of the same for 1 day. Then raise this amount to that power whose index is equal to the number of days, and it will be the amount for that time. Which amount, being multiplied by the principal sum, will produce the amount of that sum, as in the former cases.

ANNUITIES.

Annuity is a term used for any periodical income, arising from money lent, or from houses, lands, salaries, pensions, &c. payable from time to time, but mostly by annual payments.

Annuities are divided into those that are in possession, and those in reversion : the former meaning such as have already commenced ; and the latter such as will not begin till some particular event has happened, or till after soine certain time has elapsed.

When an annuity is forborne for some years, or the payments not made for that time, the annuity is said to be in arrears, or in reversion.

An annuity may also be for a certain number of years; or it may be without any limit, and then it is called a perpetuity.

The amount of an annuity, forborne for any number of years, is the sum arising from the addition of all the annuities for that number of years, together with the interest due upon each after it becomes due.

The present worth, or value, of an annuity, is the price or sum which ought to be given for it, supposing it to be bought off, or paid all at once. Let a = the annuity, pension, or yearly rent ;

n = the number of years forborne, or lent for;
R= the amount of il for 1 year;
m= the amount of the annuity;

v= its value, or its present worth. Now by compound interest, (theor. 2, we have p = Hence giving to

R

a t the successive values 1, 2, 3, n, we get

R' R?' R3

present values of a due at the end of 1, 2, 3, .. 1, years respectively. Therefore, the sum of all these will be the present value of the n years' annuities; and if n be infinite, it will be the present value of a perpetual annuity of a £ per term. Now by summing this geometrical series, we have v = 음 + + to.it

R
R?

R3
R" – 1

the present value of the annuity which is to terminate in n R” R-1' years; and if n be infinite, v= for the value of the annuity in per

R-1 petuity.

Again, because the amount of £l in n years is R", the increase in that time is R” -1; but its amount in one year, or the annuity answering to that increase

a

a

a

as the

a

a

a

a

a

a

a, and

m =

a =

-.m =

а

m

a

R“

log R

R” -1 is R-1: and as these are in the ratio of a to m, we have m =

Ꭱ the several cases relating to annuities in reversion are easily found to be as follow : R”. 1

R-1

R 1 a = vR”.... (1)

.VR" .....(4)

vR R

R” -1 R -1

m(R-1)+a

log R” -1

log m-logo (2)

(5) R R”

log R il

log m - log o (3) log. R=

(6) RP ") In theorem (3), r denotes the present value of an annuity in reversion, after p years, or not commencing till after the first p years; and it is found by taking

P
R” 1

RP 1 the difference between the two values

and

for R-1 R” R 1

RP' n years and p years. The other formulæ are derived from those in compound interest taken in connexion with the fundamental theorem deduced above.

However, for practical purposes the amount and present value of any annuity for any number of years, up to 21, will be most readily found by the two following tables. In works professedly devoted to the subject, these tables are carried to a much greater extent.

n

a

a

TABLE I.

The Amount of an Annuity of ll at Compound Interest.

1 0.9709 0.9662 0.9615 0 9569 0.9524 0.9524
2 1.9135 1.8997 1.8861 1.8727 1.8594 1.8334
3 2-8286 2.8016 2 7751 2:7490 2-7233 2.6730
4 3.7171 3:6731 3.6299 3 5875 3:5460 3:4651
5 4:5797 4.5151 4:4518 4.3900 4 3295 4.2124
6 5:4172 5.3286 5.2421 5:1579 5:0757 4.9173
7 6.2303 6:1145 6:0020 5.8927 5.7864 5.5824
8 7:0197 6.8740 7.7327 6:5959

6:4632

6:2098 9 7.7861 7.6077 7.4353 7.2688 7.1078 6.8017 10 8:5302 8.3166 8:1109 7.9127 7.7217 7.3601 11 9:5256 9:0016 8.7605 8.5889 8:3054 7.8869 12 9.9540 9.6633 9.3851 9:1186 8.8633 8.3838 13 10.6350 10:3027 9.9857 9.6829 93936 8.8527 14 11.2961 10.9205 10:5631 | 10:22 28 9.8986 9.2950 15 11.9379 11:5174 11:1184 10:7396 10:3797 9:7123 16 12:561 12:0941 11.6523 | 11.2340 10:8378 10 1059 17 13:1661 12:6513 12:1657 11.7072 11-2741 10:4773 18 13.7535 13.1897 12 6593 12:1600 11:6896 10.8276 19 14•3238 13.7098 13:1339 12:5933 | 12:0853 11:1581 20 | 14.8775 14.2124 13:5903 13.0079 12:4622 11:4699 21 15:4150 14.6980 | 14:0292 | 13:4047 | 12:3212 | 11.7641

To find the amount of any annuity forborne a certain number of years. Take the amount of il from the first table, for the proposed rate and time; then multiply it by the given annuity; and the product will be the amount, for the same number of years, and rate of interest. Also, the converse to find either the rate or the time.

Ex. To find how much an annuity of 501 will amount to in 20 years, at 3 per cent. compound interest.

On the line of 20 years, and in the column of 3} per cent. stands 28.2797, which is the amount of an annuity of 11 for the 20 years. Then 28.2797 x 50, gives 1413.9851 = 14131 19s Sd for the answer required.

To find the present value of any annuity for any number of years. Proceed here by the second table, in the same manner as above for the first table, and the present worth required will be found.

Ex. 1. To find the present value of an annuity of 501, which is to continue 20 years, at 3} per cent. By the table, the present value of il for the given rate and time, is 14.2124; therefore 14:2124 x 50 = 710:621, or 7101 12s 4d, is the present value required.

Ex. 2. To find the present value of an annuity of 201, to commence 10 years hence, and then to continue for 11 years longer, or to terminate 21 years hence, at 4 per cent. interest. In such cases as this, we have to find the difference between the present values of two equal annuities, for the two given tiines ; which, therefore, will be done by subtracting the tabular value of the one period from that of the other, and then multiplying by the given annuity. Thus, the tabular value for 21 years is 14 0292, and that for 10 years is 8:1109. Then, the difference 5:9183 multiplied by 20 gives 118.3661, or 1181 78 3d, the answer.

+

+

+

+

+

+

+

ES

+

...

1.3

1

or

+

4.6

=S

SERIES BY SUBTRACTION. This method is most readily applicable to the cases where the several terms of the series are the differences (or the same multiple of the differences) between two equi-distant corresponding terms of some other series. A few simple examples will sufficiently illustrate the practice, whilst the principle of these processes is self-evident. The only difficulty is to find the series whose differences are the terms of the given one ; and for this no general and simple rule exists.

1 1 1 Ex. 1. Let It +

+
+

ad inf. =
2 3 4
then

1
+

+

ad inf. = S- 1 3

5 1 1

1 by sub.

+ + + + ... ad inf. = 1 1.2 2.3 3.4 4.5

1 1 Ex. 2. Let it

3

+ 2

...=s.

4 1

1
then

+
5
+

6
2 2 2 2

3
by sub.

+
+

+
2.4 3.5 4.6
1 1

1
+ +

+ 1.3 2.4 3.5

1 1 1 Ex. 3. Let + + +

1.2 2.3 3.4

1 1 1 then +

+ +
2.3
4.5

2
2
2

2
by sub.

+ +
1.2.3 2.3.4

3.4.5
1
1

1
+ + +
1.2.3
2.3.4 3.4.5

1
1

1 Ex. 4. Find the sum of the series

+
2.4.6 4.6.8 6.8.10

+ Take away the last factor out of each denominator, and assume

1 1 1
+

+
2.4

4.6 6.8 1 1 1

1
then

+ +
4.6 6.8 8.10
4
4

4
+ + +
2.4.6 4.6.8 6.8.10

8
1
1

1
Hence
+ + +

Х
2.4.6 4.6.3
6.8.10

32
1 1 1
Ex. 5. Find the sum of

+

ad inf.
1.3 2.4 3.5
1 1 1 1

1
assume

+ + ...=s, then transposing
1

1
1
1

1
+
3

5

6

2 2 2 hence by subtraction

1.3

3.5
1
1
1

1 Ex. 6. Find +

+

+ ... ad inf. Ans. 2.4.6.8 4.6.8.10 6.8.10.12

2889

3.4

+

ad inf.

5.9

Ex. 9. Thus, if 2.3 +34 + ... +

1
1

1 Ex. 7. Also of

+
+

Ans.
2.5.8.11 5.8.11.14

7209 1 1 1

1 Ex. 8. Sum the series +

Ans. 1.5 3.7

6 Note. The upper line (carried to any extent) contains one term at the end, under which there is no term in the lower line. But in the above examples this circumstance creates no difference, since the terms being infinitely distant, and continually converging towards 0, that last term itself is virtually 0. However, if the law of the series be such, that the terms converge towards a limit different in value from 0, then this value being that of the last term of the upper line, must be added to the sum obtained as above.

1 . ,

+

were purposed to be summed; the

4.5
process would be, if performed according to the preceding type,

1 2 3 4
+ +

...=s; then, transposing, 2

+
3
1

1 hence by subtraction

+ +
2.3
3.4
4.5

2 But as the values of the several terms of the series, whose value is s, converge towards 1, the uncompensated term itself is 1, which added to the value already found, gives 1 – } = { for the true sum of the series. This final term, in such cases, may be called the correction of the sum. 1 1 1 1

1 Ex. 10. Find the sum of +

t... Ans.
3.5
5.7 7.9

9.11

assume

+

we have

+

+

=

+
+

MISCELLANEOUS EXAMPLES,

-.

5 7 9 11 13

5 Ex. 1. Find the sum of

+
+

Ans.
1.4 2.5 3.6 4.7 5.8

6 1 s. 1

1 1 1 Ex. 2. Show that

+
1.3 3.5 5.7

3 6 10
5
9

13
is equal to the sum of

+

+
1.2.1.3

3.4.5.7
1
1

1 Ex. 3. The sum of the series

+

+ m(m+r) (m+r) (m+2r) '(m +2r) (m+3r) 1

3 3 +11 t... ad inf.

3 Ex. 4. The sum of n terms * of a + 2ar Заr2 + 4ar3 + ... is ar" nar"

and the sum to infinity is S, = (1 — r)2 1-p

(1 — r)2: 2 3

a+b Ex. 5. Sum n terms of - +

a+26
+ + .... and of
73

t... pa

mr and likewise the sum of each series to infinity.

* This and such questions differ from the preceding in no respect but this: that the expression for the nth term must be taken as the last of the assumed series instead of the infinitely distant term towards which in that case the succeeding terms more directly point. The correction for this must be made as directed in the note above.