Page 87 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I equAtions

Proof: The system is equivalent to

2
ax +(b − 1)x 1 + c = x 2 − x 1
1
2
ax +(b − 1)x 2 + c = x 3 − x 2
2
. . .
ax 2 n−1 +(b − 1)x n−1 + c = x n − x n−1
2
ax +(b − 1)x n + c = x 1 − x n
n
2
and we can observe that Δ=(b − 1) − 4ac is the discriminant of the quadratic equation
ax +(b − 1)x + c =0 . When Δ=0 , ax +(b − 1)x i + c is a perfect square, i.e.
2
2
i
2
) ≤ 0,
ax +(b − 1)x i + c = a(x i + b−1 2 b−1 2
) (i =1, 2, 3, ··· ,n). If a< 0, then a(x i +
i
2a
2a
thus x 2 − x 1 ≤ 0,x 3 − x 2 ≤ 0, ··· ,x n − x n−1 ≤ 0,x 1 − x n ≤ 0 , which is equivalent to
x 1 ≥ x 2 ≥ x 3 ≥ ··· ≥ x n−1 ≥ x n ≥ x 1 . Hence, we can only choose equal sign in all these
inequalities, that is, x 1 = x 2 = x 3 = ··· = x n = 1−b . If a> 0, same logic follows to obtain
2a
x 1 = x 2 = x 3 = ··· = x n = 1−b . As a conclusion, the equation system has a unique solution
2a
when Δ=0.

2.97 Given f(1) = and when n> 1, f(n−1) = 2nf(n−1)+1 , find f(n).
1
5 f(n) 1−2f(n)
Solution: Multiply f(n−1) = 2nf(n−1)+1 by f(n)[1 − 2f(n)] to obtain
f(n) 1−2f(n)
f(n − 1) − 2f(n − 1)f(n)= 2nf(n − 1)f(n)+ f(n)f(n − 1) − 2f(n − 1)f(n)= 2nf(n − 1)f(n)+ f(n), which is equivalent to
f(n − 1) − f(n) = 2(n + 1)f(n)f(n − 1). Divide both sides by f(n)f(n − 1) to obtain

1 1 = 2(n + 1). Replace n with 2, 3, ··· ,n successively to obtain
f(n) − f(n−1)
1 1 =2 × 3, 1 1 =2 × 4, ··· , 1 1 = 2(n + 1)
f(2) − f(1) f(3) − f(2) f(n) − f(n−1) . Add them up to
(3+n+1)(n−1)
1 1 1 1 = 2[3+ 4+ ··· +(n + 1)] = 2 (3+n+1)(n−1) =(n − 1)(n + 4) 4)
=(n − 1)(n +
obtain f(n) − − f(1) = 2[3+ 4+ ··· +(n + 1)] = 2 × × 2 2 . Hence,
f(1)
f(n)
2
1 = 1 +(n − 1)(n +4) = 5+ n +3n − 4= n +3n +1.
2
f(n) f(1)
As a conclusion, f (n) = 1 .
n +3n+1
2
1
2.98 Solve the equation (a + a )= m .
x
−x
2

Download free eBooks at bookboon.com
87