【棒棒版】九月数学（版本号9/28 2327）、逻辑（版本号9/23 1827）寂静整理版

summer0711

thank u so much

风叶草

coralmirror

谢谢

7642

回复 马上就考了

xiazixiang

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eldertsai

数列满足：a(n)=a(n-1)-a(n-2)，a1=-1，a2=1，若n大于等于3，求前1000项的和

I wonder whether “0” is the answer?



Firstly, n≥3, so the question should ask the sum of a3~~~a1002.



a 3 =a2-a1=1+1=2

a 4=a3-a2=2-1=1

a 5=a4-a3= -1

a 6=a5-a4= -2

a7=a6-a5= -1

a8=a7-a6= 1

a9=a8-a7= 2

a10= a9-a8=1

…………..

数列为為2, 1, -1, -2, -1, 1循环

1000/6=166……4

前1000项的和=a999+a1000+a1001+a1002=a3+a4+a5+a6=2+1-1-2=0.


I wonder whether “0” is the answer?



Firstly, n≥3, so the question should ask the sum of a3~~~a1002.



a 3 =a2-a1=1+1=2

a 4=a3-a2=2-1=1

a 5=a4-a3= -1

a 6=a5-a4= -2

a7=a6-a5= -1

a8=a7-a6= 1

a9=a8-a7= 2

a10= a9-a8=1

…………..

数列为為2, 1, -1, -2, -1, 1循环

1000/6=166……4

前1000项的和=a999+a1000+a1001+a1002=a3+a4+a5+a6=2+1-1-2=0.

刘姗姗

xiexie

yche235

thanks!

sweetcat0222

Thanks a lot!!!

zitongkyd