刘占国《利息理论》第二章习题详解及提示

第二章 年金

?0.087?1??1??12??4.解：16000?A?1000?0.08712?n?120?0.087??0.087??1???1??

1212?????11??1?i??nan?x???1?i??1?xii5.解：?2n21??1?i?1??1?xi?a2n?y??(*)iid 将i?代入（*）

1?d 7.解：

1000 1000 0

1

… 1000 17

?512

18

?1000?s?1?0.08?18

?39169.84

???100.1 5000?s?Ra8.解：100.1???100.15 5000?s?Ra9.解：100.114.解：永续年金每年支付R

1?1?Ran?R??an? 2?i?17.解：1500am0.008?100000 解得m?95.6 即正常还款次数为95次 1500a950.008?f(1?0.008)?95?100000 解得f?965.74 19.解：

?1000?si(2)?si(2)?105?22105???2000si(2)?17000?5 ?2??1?i???1?i??17?1?i??15?0 令f(t)?t10?t5?17t?15

0?f(1.03)f?i?1.03(1.?03f5)(1.03)

1.0?351.032?) f(1.03?0.00 3?41??1?i??0.04?23.解：a60.04? 1.04?6，1?i??1??i4??424.解：

R 1.1025R 0

1

2

1.205R 3

4

R?1.054?1.1025R1.053?1.1025R1.052?1.205R1.05?11000 得R?2212.147

1??1?i?25.解：an?i ??n??an?i?n?1?i??n?1i?1??1?i?i2?n

?an?in?16.867i?0.1???an0.10?2an0.002 其中n通过公式（2-76）得到

0.11?v729.解: a7?i?v7?1?ia7?1?iK

类似地，v11?1?ia11?1?iL，v18?1?ia18?1?iM ?v7v11?v1831.解：a(2)n 从而i??(1?iK)(1?iL)?1? i ML?K?M KL1?vn?(2)，in??半??a(12)n1?vn?ii??(12)??(12)??an d12??i??ni?1?i半??10000 32.解：500lima21?1?i 1?i??半?1?d

?t?1半i???1?d?，i半??1?d??12?12?1

??1?i半?i半1?t?20

?95.36?a20?1?i??n?1?i?i?2036.解：?Ia?20???n?nvnai

37.解：

1 0

1

1 2

1 … 3 …

1 该永续年金现值为

i

1

0

1

2

3

1 4

5

?21 6

… …

?4

1 (2?i)i 该永续年金现值为：?1?i???1?i????11i?3? ∴所求年金现值为：?

i(2?i)i(2?i)i39.解：??1?kt?vtdt?f?g?h

0n f?liman?limn??n??1?vn??1?

1 g?(1?kn)?nv

?dr?01?r40.解：a(t)?e?1?t

t1an??a(t)dt??0n?11dt?ln(1?n) 01?tn42.解：后五年等比

??0s10???s5 100???43.解：

1

0

…

4

5

1 6

?1?k?1???1?i??10?50i?k5????i1??i1

52 7

2 8

3 9

… 10 …

3

111v4v6v8v4a8???????? ?a4??a6??iiiiii?ivd??251.022?145.解：Ks??300.015 ?5?Ka46.解：

1a300i月a120i?月1a300i月a180i1.03?10?月新xa180i1.03?10?1

月新100000dr?01?r47.解：a(t)?e?1?t

t1?1) v(0)??(t?1)a(t)dt??(t211142?1141dt?84.5 1?t48.解：at?

1?vt?，an?n1?vn?

?11?n?a??4?40 ???n?10%??n0atdt??n1?vt0?1?1?vndt??n????n49.解：1）

??Ia???t?1tt?1??t?tvtai

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