two math question

同學請留意一個常見的技巧：
　　　　　(a-b) = -(b-a)

A. Simplify 2m/(m-2)+m/(2-m).

2m/(m-2)+m/(2-m)

= 2m/(m-2)-m/(m-2)　<-- Use the technique

= (2m-m)/(m-2)

= m/(m-2)


注意： 1+2+3+...+n = (1+n)(n)/2


B. The general term of a sequence is 5n-4. find the sum of the first 10 terms of the sequence.

If I denote the k-th term as T(k), then T(n) = 5n-4.
The question asks to find T(1)+T(2)+...+T(10).

Note that
T(1) = 5(1)-4
T(2) = 5(2)-4
T(3) = 5(3)-4
T(4) = 5(4)-4
T(5) = 5(5)-4
T(6) = 5(6)-4
T(7) = 5(7)-4
T(8) = 5(8)-4
T(9) = 5(9)-4
T(10) = 5(10)-4

Adding these lines gives
5(1+2+3+...+10) - 4(10)

= 5*(1+10)*10/2 - 40

= 5*55-40

= 275-40

= 235

2013-08-30 17:26:46 補充：
原理是這樣的：

設　Ｓ　＝　１＋　　２　　＋　　３　　＋．．．＋（ｎ－１）＋ｎ

同時Ｓ　＝　ｎ＋（ｎ－１）＋（ｎ－２）＋．．．＋　　２　　＋１

那麼
　２Ｓ　＝　（ｎ＋１）＋（ｎ＋１）＋．．．＋（ｎ＋１＿）
　　　　＝　ｎ（ｎ＋１）

所以
　　Ｓ　＝　ｎ（ｎ＋１）／２

岩岩教呢課,但未教sum of first nth terms,
書中搵到:
the sum of first nth terms=(number of terms/2)(first term+the nth term)