By C. W. Celia, A. T. F. Nice, K. F. Elliott

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**Extra resources for Advanced mathematics 1**

**Sample text**

J q Algebra 1 2& product of the roots = 11 1 --fJ =(X q the required equation is x 2 - (- ~ )x + ~ = 0, qx 2 + px + 1 = 0. e. e. 1 +py+qy 2 = o, where y = ljx, and so the roots are the reciprocals of the roots of the given equation. The relations between the roots and the coefficients of equations of higher degree can be found. Let IX 1, IX 2, tX 3 be the roots of the cubic equation ax 3 + bx 2 +ex+ d = 0. e. a(x- IXd (x- IX 2)(x- 1X 3) = 0, a[ X 3 - (lXI + IX2 + 1X3)X 2 + (lXI IX2 + IX2IX3 + IX31Xdx- IX1 1X21X3] = 0.

N(n + 1) = r, the sum of the first n natural numbers. L r=1 Algebra 2 &1 The series 1 + 2 + 3 + . . + n is an example of an arithmetic progression, the general form of which is a+(a+d)+(a+2d)+ ... +(1-d)+l. where a is the first term, I is the last term and d is the common difference between successive terms. Again and S" = sn = Adding 2S" a +(a+d)+(a+2d)+ ... +(1-d)+ + (1- d)+ (I - 2d) + ... +(a+ d)+ a = (a+ I)+ (a+ /)+(a+ /) + ... n(a +I). n[2a + (n -1)d]. e. n(n + 1)- n n giving r L r = 1 Example n L Evaluate r r(r + 1).

Or 0 _~ 1 < 2" 20 ()()() < 2 n log1 0 20000 < n 1og10 2 4·3010 n > 0·3010 (~ 14·29). Thus the least number of terms required is 15. Example 4 A man borrows £10000 on 1st January 1979. Compound interest at the rate of 12 per cent is added on the 31st December each year. The man is to repay the loan in twenty equal annual instalments of £x on 1st January of each year, the first repayment being made on 1st January 1980. Calculate the value ofx to the nearest integer. e. e. e. £10000 x 1·123 - £x(1 + 1·12 + 1-122) (after three repayments).