奇數(shù)項(xiàng)共n+1項(xiàng)，其和為S奇=
| (n+1)(a1+a2n+1) |
| 2 |
| (n+1)2an+1 |
| 2 |
偶數(shù)項(xiàng)共n項(xiàng)，其和為S偶=
| n(a2+a2n) |
| 2 |
| n2an+1 |
| 2 |
| ① |
| ② |
| n+1 |
| n |
| 4 |
| 3 |
故選A
等差數(shù)列{an}共有2n+1項(xiàng)，其中奇數(shù)項(xiàng)之和為4，偶數(shù)項(xiàng)之和為3，則n的值是（ ?。?A.3 B.5 C.7 D.9
等差數(shù)列{an}共有2n+1項(xiàng)，其中奇數(shù)項(xiàng)之和為4，偶數(shù)項(xiàng)之和為3，則n的值是（ ?。?br/>A. 3
B. 5
C. 7
D. 9
B. 5
C. 7
D. 9
數(shù)學(xué)人氣：823 ℃時(shí)間：2020-06-23 13:55:25
優(yōu)質(zhì)解答
設(shè)數(shù)列公差為d，首項(xiàng)為a1，
奇數(shù)項(xiàng)共n+1項(xiàng)，其和為S奇=
=
=（n+1）an+1=4，①
偶數(shù)項(xiàng)共n項(xiàng)，其和為S偶=
=
=nan+1=3，②
得，
＝
，解得n=3
故選A
奇數(shù)項(xiàng)共n+1項(xiàng)，其和為S奇=
偶數(shù)項(xiàng)共n項(xiàng)，其和為S偶=
故選A
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