數列與函數一Ｄ問題

A) sequence follow by 1^2 , 2^2 , 3^2 ...... n^2 so the 5th term is 5^2 = 25

B) follow by 1^2+1, 2^2+1, 3^2+1.... n^2+1 so the 4th term is 4^2+1 = 17

C) follow 2(1)+1, 2(2)+1 ,3(2) +1... 2(n)+1 so 5th term is 2(5) +1 = 11

D) follow 4(1)+2, 4(2)+2, 4(3)+2 ... 4(n)+2 so 3rd term is 4(3)+2 = 14

E) follow 1/(1+1) , 2/(2+1), 3/(3+1) ... n/(n+1) so the 5th term is 5/(5+1) = 5/6

Ａ）

漏空既項= 25 ( 因為1=1^2, 4=2^2, 9=3^2, ...)
通項= n^2 (即係n既二次方)

Ｂ）

漏空既項= 17
通項= (n^2)+1 (因為全部都係 a part 既答案再加1)

Ｃ）

漏空既項= 11
通項= 3+2(n-1) (因為 3+(2)=5, 5+(2)=7, 7+(2)=9, ...)

Ｄ）　

漏空既項= 14
通項= 6+4(n-1) (因為逐項加4)

Ｅ）

漏空既項= 5/6
通項= n/(n+1)