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# 高二数学等比数列前n项和5

1.等比数列的定义是什么? 等比数列的定义是什么?
an = q ( n ≥ 2) an 1

a1 , a1q , a1q , L , a1q

2

n 1

,L

S30
(1 + 30) 30 × = 1 + 2 + 3 + L + 30 = =465(万元 =465(万元) 万元) 2
,则

2 3

T30 = 1 + 2 + 2 + 2 + L + 2 + 2
28 2

29

2T30 + 2 +L 2 + 2 2T30 = 2 + 2 + 2 + 2 + L + 2 + 2 + 2
23 3
28

28 29

29 30

30

T30 = (2 1)分
30

S n = a1 + a1q + a1q + L + a1q
2

n 1

n

a1 (1 q n ) 1 q Sn = na 1 q ≠1 q =1

a1 (1 q ) Sn = (q ≠ 1) 1 q

1 1 1 1. 求等比数列1, , , , L 前10项的和. 2 4 8 an 2. = 5 + 5 + L + 5(100个5的和) S100

= a1q

n 1

a1 (1 q n ) Sn = 1 q na 1

q ≠1 q =1

3,判断正误:

1× (1 2 20 ) 1 + 2 + 2 2 + 23 + 2 4 + L + 2 20 = 1 2

4, 等比数列{an }中, a1 = 6, q = 2, an = 192,

a1 (1 q n ) a1 qan q ≠1 = 1 q 1 q Sn = q =1 na1

= 1 + 2(1 + 2 + 2
29

2

+ 23 + L + 2 28 )

= 1 + 2(T30 2 )

T30 = 2 1
30

S n = a1 + a1q + L + a1q

n2

+ a1q

n 1

a1 (1 q n ) = 1 q

q ≠1

S n = a1 + a1q + a1q + L + a1q
2

n2

+ a1q

n 1

= a1 + q (a1 + a1q + a1q 2 + L + a1q n 2 )

∴ (1 q) S n = a1 a1q

= a1 + q( S n a1q

n 1

)
n

1 1 1 1 求和 : S n = 1 + 2 + 3 + L + ( n + n ) 2 4 8 2 1 1 1 1 = (1 + ) + (2 + ) + (3 + ) + L + (n + n ) 4 8 2 2
(1 n ) n(n + 1) 2 2 = + 1 2 1 2

1 1 1 1 = (1 + 2 + 3 + L + n) + ( + + + L + n ) 2 4 8 2 1 1

n(n + 1) 1 = n +1 2 2

T30 = 1 + 2 + 2 2 + 23 + L + 2 28 + 2 29

a1 (1 q n ) q ≠1 1 q Sn = na q =1 1

1,书本30页: 第8题的偶数题; 第10题 2,探求:等比数列前n项和公式的其它证明方法.