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2.3 等差数列的前 n 项和 (2)

2.3 等差数列的前 n 项和 (2). 复习 : 数列 前 n 项和 :. 性质 ：若数列 前 n 项和为 ，则. 等差数列的前 n 项和公式 ：. 或. 等差数列的前 项和. 两个公式都表明要求 必须已知 中三个. 注意：. 补充例题. 如图，一个堆放铅笔的 V 形架的最下面一层放 1 支铅笔，往上每一层都比它下面一层多放 1 支，最上面一层放 120 支 . 这个 V 形架上共放了多少支铅笔？. 解：由题意知，这个 V 型架自下而上各层的铅笔数成等差数列，记为｛ a n ｝.

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2.3 等差数列的前 n 项和 (2)

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  1. 2.3等差数列的前n项和(2)

  2. 复习: 数列 前n项和: 性质：若数列 前n项和为 ，则

  3. 等差数列的前n项和公式： 或 等差数列的前 项和 两个公式都表明要求 必须已知 中三个 注意：

  4. 补充例题 如图，一个堆放铅笔的V形架的最下面一层放1支铅笔，往上每一层都比它下面一层多放1支，最上面一层放120支. 这个V形架上共放了多少支铅笔？ 解：由题意知，这个V型架自下而上各层的铅笔数成等差数列，记为｛an｝. 答：V型架上共放着7260支铅笔。

  5. 课本Ｐ51. 探究: 若一个数列 　的前n项和为Sn=pn2+qn+r,其中p、q、r为常数，且p　０ ，那么这个数列一定是等差数列吗？若是， 它的首项与公差分别是什么？ (解答过程在下一个幻灯片) ｛an｝

  6. 解: ∵当n=1时, a1=s1=p+q+r. 当n≧2时, an= sn-sn-1= =2pn-p+q p+q+r (n=1) • ∴ an= 2pn-p+q (n≧2) 当r=0时,an=2pn-p+q(n≧1) 是等差数列,首项是a1=p+q 公差是d=2p.

  7. 例4．己知等差数列 5, 4 , 3 , … 的前n项和为Sn, 求使得Sn最大的序号n的值. 解:由题意知,等差数列5, 4 , 3 , …的公差 为 ,所以sn= [2×5+(n-1)( )] = = ( n- )2+ 答案: (7或8).

  8. 联系: an = a1+(n-1)d的图象是相 应直线 上 • 一群孤立的点.它的最值又是怎样?

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