問2條F.1 Maths問題

16) (-101)+102+(-103)+104+...+(-199)+200

=(102-101)+(104-103)+...+(200-199)

=99x1

=99

17a)1st term=2^2+2

2nd term=3^2+2

3rd term=4^2+2

4st term=5^2+2=27

5nd term=6^2+2=38

6rd term=7^2+2=51

17b)The nth term=(n+1)^2+2

The 10th pattern=(10+1)^2+2=123

17c)
圖片參考：http://imgcld.yimg.com/8/n/HA04758875/o/701107220089713873431590.jpg

The nth pattern=n+4n=5n

The 29th pattern=145

2011-07-22 20:26:22 補充：
Should be:

17c)

(picture)

2011-07-22 20:29:22 補充：
Sorry,reading mistake:

Let the nth term be x

x=(n+1)^2+2

145=(n+1)^2+2

(n+1)^2=143

There is no such pattern

(-101)+102+(-103)+104+...+(-199)+200

= (-101 + 102) + (-103 + 104) + ... + (-199 + 200)

= 100

17(a)

T_4 = 5*4 + 7 = 27

T_5 = 6*5 + 8 = 38

T_6 = 7*6 + 9 = 51

(b) T_10 = 11*10 + 13 = 110 + 13 = 123

(c) T_11 = 12*11 + 14 = 132 + 14 = 146

So, there is no such pattern