Page 28 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution: According to the rule in the terms of S, we obtain the general formula

1
1
1
1
1 2
) − 2
a n = 1+ n 2 + (n+1) 2 = (1 + ) − 2 + (n+1) 2 = ( n+1 2 n+1 1 + (n+1) 2 =

n
n n+1 n+1
) − 2
n 1 2
a n = 1+ 1 2 + 1 2 = n (1 + ) − 2 + 1 2 = ( n+1 2 1 + 1 2 =
n (n+1) n n (n+1) n n n+1 (n+1)
( n+1 − 1 ) = n+1 − 1 = 1+ 1 − 1
2

n n+1 n+1 1 2 n n+1 1 n n+1 1 1 . Thus
n+1
( − ) = − = 1+ −
n n+1 n n+1 n n+1
1
1
1
1
1
1
1
1
1
1
S = (1+ − )+(1+ − )+(1+ − )+···+(1+ 2009 − 2010 )+(1+ 2010 − 2011 ) = 2010 2010

1
4
3
3
2
2
1 2010 2011
2011 − 2011 = 2010 2011 ∈ (2010, 2011)
∈ (2010, 2011) which implies that S has the integer part 2010.
3
√ x + x +1
1.75 Given x = 5+1 , evaluate .
2 5
x
√
Solution: Let y = 5−1 , then xy =1,x − y =1.
2
2
3
2
3
x + x +1 x + x + xy x +1+ y x + x − y + y x +1 x + xy
3
2
2
3
x + x +1 x + x + xy x +1+ y = x + x − y + y = x +1 = x + xy =
=
=
3
3
5
x 5 x 5 = x √ 5 = x 4 x 4 = x 4 x 4 = x 3 = x 3 =
x
x
1+ y = x = 1 = y = x √ 5 − 1
5 − 1
x
1
1+ y
=
x 2 x 2 x 2 = x 2 x x = y = 2 2 .
1.76 M is a 2000-digit number and a multiple of 9. M 1 is the sum of all digits
of M , M 2 is the sum of all digits of M 1, and M 3 is the sum of all digits of M 2. Find the
value of M 3.
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