¿©·¯ °¡Áö ¼ö¿­ÀÇ ÇÕ

1. µîÂ÷, µîºñ¼ö¿­ÀÇ ÇÕ ¡æ °ø½ÄÀ» ÀÌ¿ëÇÕ´Ï´Ù.^^

1) µîºñ¼ö¿­ÀÇ ÇÕ : Sn = (n/2)(a+l) = (n/2){2a+(n-1)d}
2) µîÂ÷¼ö¿­ÀÇ ÇÕ : Sn = a(rn-1)/(r-1) = a(1-rn)/(1-r) (r¡Á1)

2. ±âŸ ¼ö¿­ÀÇ ÇÕ : Sn = a1+a2+a3+¡¦+an =ak ¸¦ ÀÌ¿ë. ¡ç an = a1+bk (n = 2,3,4, ¡¦)

3. ¸è±Þ¼ö : S-xS = (µîºñ¼ö¿­ÀÇ ÇÕ) + ¥á ¸¦ ÀÌ¿ë

4. ºÐ¼ö¼ö¿­, ±ÙÈ£°¡ ÀÖ´Â ¼ö¿­ ÀÇ ÇÕ : ÀÌÇ׺и®ÇÏ¿© ÇÕÀ» ±¸ÇÑ´Ù.


3)

Problem 3-5 ¡æ ¹®Á¦¸¦ ´©¸£¸é Ç®ÀÌ¿Í ´äÀÌ ³ª¿É´Ï´Ù.

  1. ¼ö¿­ 12, 42, 72, 102, 132, ¡¦ ÀÇ Á¦n Ç×±îÁöÀÇ ÇÕ Sn À» ±¸ÇϽÿÀ.
  2. (´ä) (1/2)n(6n2-3n-1)

    1) ¼ö¿­ 1, 4, 7, 10, ¡¦Àº ù° Ç× 1, °øÂ÷ 3 ÀÎ µîÂ÷¼ö¿­ ¡æ n° Ç×Àº 1+(n-1)¡¤3 = 3n-1
    2) ¼ö¿­ 12, 42, 72, 102, 132, ¡¦ ÀÇ ÀϹÝÇ× ¡æ an = (3n-2)2
    3) Sn =ak ¿¡¼­ Sn =(3k-2)2 =(9k2-12k+4) = 9k2 -12k +4n
    4) Sn = 9(1/6)n(n+1)(2n+1) - 12(1/2)n(n+1) +4n = (3/2)(2n3+3n2+n)-6(n2+n)+4n
    5) Sn = 3n3-(3/2)n2-(1/2) = (1/2)n(6n2-3n-1) ^^

    ¢Ñ k = 1+2+3+ ¡¦ +n = (1/2)n(n+1), k2 = 12+22+32+ ¡¦ +n2 = (1/6)n(n+1)(2n+1)

  3. S =(Á¦ 99Ç×) ÀÏ ¶§, S ¸¦ ±¸ÇϽÿÀ.¡¡
  4. (´ä) 9

    ¢Ñ



  5. ¼ö¿­ 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, ¡¦  ÀÇ Á¦10 Ç×±îÁöÀÇ ÇÕÀ» ±¸ÇϽÿÀ.
  6. (´ä) 220

    1) Á¦k Ç× ak = 1+2+3+ ¡¦ + k = (1/2)k(k+1)
    2) S10 = (1/2)k(k+1) = (1/2)¡¿(1/3)¡¿10¡¿11¡¿12 = 220 ^^

    ¢Ñk(k+1) = (1/3)n(n+1)(n+2), k(k+1)(k+2) = (1/4)n(n+1)(n+2)(n+3)



  7. S = 1¡¤10 + 3¡¤9 + 5¡¤8 +7¡¤7 + ¡¦ + 19¡¤1 ÀÏ ¶§, S ¸¦ ±¸ÇϽÿÀ.
  8. (´ä) 385

    1) 1, 3, 5, 7, ¡¦ , 19 ÀÇ Á¦k Ç× = 1+(k-1)¡¤2 = 2k-1 À̰í 19 ´Â Á¦10 Ç×
    2) 10, 9, 8. 7, ¡¦ ÀÇ Á¦k Ç× = 10+(k-1)¡¤(-1) = -k+11
    3) 1¡¤10, 3¡¤9, 5¡¤8, 7¡¤7, ¡¦, 19¡¤1 ÀÇ Á¦k Ç× ak = (2k-1)(-k+11) = -2k2+23k-11
    4) S =(-2k2+23k-11) = -2k2 +23k - 110
    5) S = -2(1/6)¡¿10¡¿11¡¿21 + 23¡¿(1/2)¡¿10¡¿11 - 110 = -770+1265-110 = 385 ^^

    ¢Ñ a1b1 + a2b2 + a3b3 + ¡¦ + anbn = akbk



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