when n =

25, this fraction is found to be 41.63.

r.04 is too small.

therefore

Suppose, moreover. that r =. 045; then for n = 25, it follows that

A

P

Since the result is greater than (2), the number r=.045 is too large. Suppose, further, that .044; then since n = 25, it follows that

the number r = .044 is still too large. Suppose r = .043. Then it follows that

(1+r)”—1 (5)

43.37 ...

therefore r = .043 is too small. It follows from (4) and (5) that .043 < r < .044

.043+.044
2

and = .0435 is the value of r with an error less than .0005; the rate is therefore 4.35%, with a possible error of .05 of 1%. On continuing this process, as close an approximation as desired may be found.

REFUNDING OF A DEBT BY ANNUITIES

596. Sinking Fund. To make the calculations for a sinking fund is to calculate the purchase price of a given annuity.

Suppose that B desires, by paying down at once a sum of $E, to secure for himself and his heirs the right of receiving n annual payments of $P each, the first payment to be made m years hence. E is the sum of the present values of the n payments. The first payment is due in m years hence; its present value is, therefore, Р The second is due in m +1 years hence; therefore the present value of $P is and so on. Hence

(1+r)m

P

(1 + r)m+1 '

COROLLARY. It the annuity is not "deferred," but begins at once, i. e., the first payment is due in one year, then m = 1, and

597. The Calculation of E or of P.-These two problems do not cause any difficulty. If the unknown is E, then from (3), 1596, (4) log E = log P + log [(1 + r)" — 1] — n log (1 + r) — log r. Application. What is the value of a loan E which can be refunded by 34 annuities of $1500 each, the rate being 41% and the first annuity being paid at the end of one year?

=

34; using logarithms one finds

(1+r)"4.46626;

log E log 1500+ log (3.46626)-n log (1.045)-log .045 = 3.176091 +0.539862.649944-8.653213 × 10

= 4.412796;

E$25870.

598. If P is unknown it can be found by solving equation (4), 2597, for log P, etc.

599. To find n, solve equation (3) 596, for (1+r)", giving

and determiner by successive approximations, employing the method of 2595.

601. The Period of Time a Fraction of a Year.-Suppose that the interest is compounded every six months instead of every year (as was the case in 1597), and in this case suppose that P' is the sum paid each period of six months, the interest on $1 for six months, and n' the number of half-yearly payments; then it follows that (1 + r′)n' — 1 ̧

P

A P

[2589, (i)]

But here P' is is and n' is 2; hence the compounding formula is

A = P

r

A similar formula holds for payments made at the end of other fractional parts of a year.

PROBLEMS

1. Find the amount of $100 in 50 years at 5% interest, compounded annually.

2. In how many years will a sum of money double itself at 41% interest, compounded annually?

3. If in the year 1776 $1000 had been left to accumulate for 124 years, find the amount in the year 1900, reckoning compound interest at the rate of 5% per annum.

4. If a sum of money doubles itself in 40 years at simple interest, find the rate of interest.

5. Find the present value of $10000 due in 10 years hence at 4% interest, compounded annually.

6. Find the amount of an annuity of $100 in 15 years, allowing compound interest at 4% per annum.

7. What is the present value of an annuity of $1000 due in 30 years, allowing compound interest at 5% per annum?

8. What sum of money at 5% interest, compounded annually, will amount to $1000 in 16 years?

9. In how many years will a sum of money treble itself at 34% interest, compounded annually?

10. A person borrows $1225 to be repaid in 5 years by annual installments of $220; find the rate of interest if simple interest is allowed on the payments.

11. A person borrows $60025; find how much he must pay annually that the whole debt may be discharged in 35 years, allowing simple interest at 4%.

12. A merchant marks his goods with two prices, one for ready money and the other for a credit of 6 months; find the ratio the two prices ought to bear to one another, allowing 5% simple interest.

13. Find the amount of an annuity of $200 in 25 years at 41%, compound interest.

14. A county treasurer borrows $50000, and wishes to repay it in 25 annual payments, the first of which should be paid one year after the loan was made: rate 4% compound interest: what ought the amount of each annuity to be? (Compare 2598.) Ans. $3200.59.

15. How often should one pay $8869.90 in order to refund a debt of $100000, the first annuity being paid one year after the debt was contracted and the rate being 5% interest, compounded annually? (Compare 2599.)

Ans. 17.

BOOK VI

CHAPTER I

MATHEMATICAL INDUCTION

602. In simple cases already considered, but more particularly in subsequent parts of this work, it is convenient to use a method of proof which is called Mathematical Induction. This method will

First Step.-The following theorem holds in case of equations (1), (2), (3), (4):

THEOREM. -If a number of the first consecutive integers are added, their sum is one-half of the number of integers added times the number of integers plus 1.

Second Step.-Assume that this law holds for the sum of the first n integers, then

(5)

1+2+3+ . . . . + n = " ( n + 1 ) ̧

Add n + 1 to both members of this equation,

2

n(n+1)

=

+ n + 1

The same law is expressed in (6) that is expressed in (5); i. e., if the theorem holds for n integers, it holds for n + 1 integers.

Third Step.-But in (4) it is noted that the theorem is true when n = 5, therefore it follows from the Second Step that the theorem holds for n = 6, then for n = 7, and so on. n integers has been found in 1526.

604. Sum the First n Odd Integers.

Suppose it is observed that

The sum of the first

First Step.-For these four cases, one sees that the following theorem holds:

THEOREM. The sum of a number of the first odd integers is the square of the number of integers added.

Second Step.-Assume that this theorem holds for n odd integers,

Add the (n+1)th odd integer, 2n + 1, to each member of (5), (6) 1+3+5 +

+2n-1+2n + 1 = n2 + 2n+1 = ( n + 1 )2.

It follows from equations (5) and (6), that if the theorem holds for the sum of n odd integers, it holds for the sum of n + 1 odd integers.

Third Step.-But the theorem holds when n = 4, therefore it holds when n = 5, hence when n = 6, and so on.

605. Suppose that one desires to prove the formula,

This formula is true in the case n = 2, 3, 4; thus,

6

but it is desired to show that the formula holds universally. Suppose that formula (i) is true for any number of terms, say r; then

r(r+1) (2r+1)