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k =1
k =1
(i)若
a
n
bn
=
試求 2n+3
A
5
3n+4
B5
=?
解:
A
5
=
(
a1+a5
)×
5 2
=
a1+a5
=
2a3
=
a3
=
2×3+3 =
9
B5
(b1+b5
)×
5 2
b1+b5
2b3
b3
3×3+4 13
(ii)若
An Bn
=
2n+3 3n+4
試求
a5 b5
=?
解:
a
5
=
2a5
=
a1+a9
=
(a1+a9
階差數列與等差數列的性質推廣
第一頁
(一)數列 an 看不出其規律,, an+1 − an = bn 為數列 an 的第一階差數列,,若 bn 是有明顯規律的數列
n −1
n −1
n −1
n
∑ ∑ ∑ ∑ 則利用 bk = (ak+1 − ak ) = an − a1 ⇒ an = a1 + bk 便可求出 an ,甚至求出級數和 Sn = ak
=
4
⇒
a1
+
a10
=
8 10
………..(1)
S20
= 12
⇒
(a1 +a20 )×20 2
= 12
⇒
a1
+
a20
=
12 10
……..(2)
由(2)-(1)知10
D
=
4 10
S30
=
(a1 +a30 )×30 2
= 15(a1
+
a30 )
= 15(a1
+
a20
+ 10 D)
=
15(
12 10
+
4 10
=
n(n
+ 1)(
2 n +1 3
−
1 2
)
=
n(n
+ 1)
4 n −1 6
例題二： an+1
= an
+ n2
− n +1且 a1
= 1且 an
=
f
(n)i
n 3
求
f
(n) = ?
解：由 an+1 − an = n2 − n +1 = n(n −1) +1
⇒ bn = an+1 − an = n2 − n +1 = n(n −1) +1 是有規律的數列
k =1
k =1
k =1
k =1
例題一：求數列1, 6,15, 28, 45, 66,91,190,....... 之第 n 項及前 n 項的和
解： an → 1, 6,15, 28, 45, 66,91,190,....... 看不出規律
bn → 5,9,13,17, 21, 25, 29,..... 明顯看出是等差數列
n−1
n −1
n −1
∑ ∑ ∑ ⇒ bk = (ak+1 − ak ) = an − a1 ⇒ an = a1 + bk
k =1
k =1
k =1
n−1
n −1
n−2
∑ ∑ ∑ ⇒ an = a1 +
bk = 1+
(
k (k −1) 2
+ 1)
=
1+
(n
−1)
+
= n + k (k +1) 2
( n − 2 )( n −1) n 3
n −1
n−1
n −1
k −1
∑ ∑ ∑ ∑ 由 an = a1 + bk 且 bn = b1 + ck ⇒ an = a1 + (b1 + ci )
k =1
k =1
k =1
i=1
例題三：求數列 2,10,30, 68,130, 222,....... 之第 n 項及前 n 項的和
解： an → 2,10,30, 68,130, 222,....... 看不出規律
=
2n(n +1)
ball)
例題四： a1 = 1,, a2 = 3,, a3 = 7, , a4 = 13, a5 = 21,,.......... 即 ak − ak−1 = f (k) − f (k −1) = 2(k −1)
n
n
n
n−1`
⇒ ∑[ f (k) − f (k −1)] = ∑ 2(k −1) ⇒ f (n) − f (1) = 2∑ (k −1) = 2∑ k = n(n −1)
k =1
k =2
n −1
n−1
n−1
n −1
由 an = a1 + ∑ bk = 2 + ∑ (3k 2 + 3k + 2) = 2 + 3∑ k(k +1) + ∑ 2
k =1
k =1
k =1
k =1
=
2
+
3×
( n −1) n ( n +1) 3
+
2(n
−1)
=
n3
−
n
+
2n
=
n3
+
n
n
n
n
∑ ∑ ∑ 韻律操公式：(1)
S2n − Sn = an+1 + an+2 + an+3 + ......... + an+n
S3n − S2n = a2n+1 + a2n+2 + a2n+3 + ......... + a2n+n
S4n − S3n = a3n+1 + a3n+2 + a3n+3 + ......... + a3n+n
n −1
n−1
∑ ∑ 由 an = a1 +
bk = 1+
(4k
+ 1)
=1+
(n
−1)
+
4i
(
n −1) 2
n
=
2n2
−
n
k =1
k =1
n
n
n
n
∑ ∑ ∑ ∑ Sn =
an =
(2k2 − k) = 2 k2 −
k = 2 − n(n+1)(2n+1) n(n+1)
6
2
k =1
k =1
k =1
k =1
由(1)+(2)+(3)+(4)+(5)得
S50
=
5S10
+
(1 +
2
+
3+
4)d
=
20
+
4×5 2
×
4
=
60
得到
Sn×10
=
n × S10
+
(1 +
2+3+
4
+ .... +
n
−1)D
=
n × S10
+
( n −1) n 2
×4
=
2n
+
2n2
=
2n(n
+ 1)
解三：
S10
=
4
⇒
(a1 +a10 )×10 2
bn → 8, 20, 38, 62,92,..... 看不出規律
cn → 12,18, 24,30,.....明顯看出是等差數列
n−1
n−1
n
∑ ∑ ∑ 由 bn = b1 +
ck = 8 +
6(k +1) = 8 + 6
k
=8 + 6×
( n −1)( n + 2 ) 2
=
3n2
+ 3n
+
2
k =1
k
=
k (k +1) 2
(2)
k(k
+ 1)
=
k (k +1)(k +2) 3
,,
k (k
+ 1)(k
+
2)
=
k (k +1)(k +2)(k +3) 4
k =1
k =1
k =1
階差數列與等差數列的性質推廣
第二頁
(三) a1, a2 , a3 成等差則 a1 + a3 = a2 + a2 = 2a2 (四)m,n,p,q∈ N 且 m + n = p + q 則 am + an = ap + aq
k=2
求(1) an = n(n −1) +1
k =2
k=2
k =1
n
n
n −1
∑ ∑ ∑ (2)
ak =
[k(k −1) +1] = n +
[k(k
+ 1)]
=
n+
