GWD5-8

[ted186]

Q8:
The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

這是數列題嗎?該如何解

[huacy]

a[sub]1[/sub] = 2
a[sub]2[/sub] = -3
a[sub]3[/sub] = 5
a[sub]4[/sub] = -1
a[sub]n[/sub] = a[sub]n-4[/sub]
a[sub]97[/sub]=a[sub]93[/sub]=a[sub]89[/sub]=.....=a[sub]1[/sub]

a[sub]1[/sub]+a[sub]2[/sub]+a[sub]3[/sub]+a[sub]4[/sub]+....+a[sub]96[/sub]=(a[sub]1[/sub]+a[sub]2[/sub]+a[sub]3[/sub]+a[sub]4[/sub])*(96/4)=72
72+a[sub]97[/sub]=72+2=74

如果可以 發文時最好連答案也一起附上 最少方便判斷 您那邊有問題

[ted186]

真不好意思~~乍到此地
先前知道該附上題目
卻忽略了答案~~
還有有些題目可能是以前數學不好
或是忘記了~~都不知道給他從何問起說
歹謝 :-$

[zachary]

a[sub]1[/sub] = 2
a[sub]2[/sub] = -3
a[sub]3[/sub] = 5
a[sub]4[/sub] = -1
a[sub]n[/sub] = a[sub]n-4[/sub]
a[sub]97[/sub]=a[sub]93[/sub]=a[sub]89[/sub]=.....=a[sub]1[/sub]

a[sub]1[/sub]+a[sub]2[/sub]+a[sub]3[/sub]+a[sub]4[/sub]+....+a[sub]96[/sub]=(a[sub]1[/sub]+a[sub]2[/sub]+a[sub]3[/sub]+a[sub]4[/sub])*(96/4)=72
72+a[sub]97[/sub]=72+2=74

如果可以 發文時最好連答案也一起附上 最少方便判斷 您那邊有問題

還是不太懂...如何知道a97=a1? (不會是一個一個遞減算出的吧..)
還有，a1+a2+a3+a4...+a96=(a1+a2+a3+a4)*(96/4)是怎麼來的呢？.....

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sorry..我自己想出了why..。我想我問了一個蠢問題...。不好意思～

[pastatasty]

Q8:
The infinite sequence a1, a2,…, an,… is such that a1 = 2, a2 = -3, a3 = 5, a4 = -1, and an = an-4 for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

a1=2
a2=-3
a3=5
a4=-1
a5=a1=2
a6=a2=-3
a7=a3=5
a8=a4=-1
...以上推出4個為一循環
a1+...+a97=(a1+...+a4)+(a1+...+a4)...共24個循環[/color[color=red]]（97=4*24+1)...+a1=24[2+(-3)+5+(-1)]+2=74