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 31% 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 598.) 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 now be illustrated. 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 Add n + 1 to both members of this equation, 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. The sum of the first n integers has been found in 1526. = 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 It follows from equations (5) and (6), that if the theorem holds for the sum of odd integers, it holds for the sum of n + 1 odd integers. Third Step. holds when n = But the theorem holds when n = 4, therefore it 5, hence when n = 6, and so on. 605. Suppose that one desires to prove the formula, (i) + n2 = n ( n + 1 ) ( 2r +1) ̧ 6 1o +22+32 + 4o + This formula is true in the case n = 2, 3, 4; thus, 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 (1). 1 + 22 + 32 + . . . . + p2 = r ( r + 1) (2r+1). 1+22 6 Add (r+ 1) to both members of this equation, then (2) (r+1) (r+2 ) ( 2r+3) = = 6 (+1)[(+1)+1][2(r+1)+1]. 6 Thus we see that (+1) is involved in the same manner in the second member of equation (2) asr is in the second member of equation (1). That is, if formula (i) holds for any number of terms, whatever that number may be, it holds when the number is increased by one. But the formula does hold by actual calculation when 2, 3, or 4 terms are taken, therefore it holds when 5 terms are taken, and Hence the formula must hold universally. so on. 606. The three theorems which have been proved by the method of induction may be established otherwise. The first and second theorems are examples in A. P., and have been proved in 1526. There are many other theorems which can be proved readily by induction. The theorem proved in 102, respecting the divisibility of x"±a" by x±a, may be proved by induction. For example, hence "a" is divisible by x- -a, when x2-1 - an-1 is. But a, - a, therefore x2. a2 is divisible by x "a" is always divisible by x- - a when n is a x -a is divisible by x and so on; hence positive integer. Similarly, other cases may be proved. As another exercise the student may consider the theorem in 268. 607. Proof of the Binomial Theorem for Positive Exponents.As a last example illustrating the method of mathematical induction, a proof of the Binomial Theorem, for a positive integral exponent, stated in 265, is here given. First Step.-It was proved in $265 that the Binomial Theorem holds in the particular cases: Second Step.-It is assumed that the theorem holds for any positive integral exponent », then Multiply both members of (2) by x+a, then, Equation (4) shows that if the binomial formula in (2) holds for the exponent n, the binomial formula holds for the exponent n + 1. Third Step.-Therefore, on combining the results of the first and second steps, the binomial formula in (1) is true for any positive integral exponent, because if it holds for any exponent, it holds for an exponent one greater; but it holds for the exponent 4, hence for the exponent 5, then for 6, and so on for any exponent. * The (r2)th term of the expansion (4) is found as follows: = xn-rar+1. |