¿©·¯ °¡Áö ¼ö¿ÀÇ ÇÕ
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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.

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Problem 3-5 ¡æ ¹®Á¦¸¦ ´©¸£¸é Ç®ÀÌ¿Í ´äÀÌ
³ª¿É´Ï´Ù.
- ¼ö¿ 12, 42, 72, 102, 132, ¡¦ ÀÇ Á¦n Ç×±îÁöÀÇ ÇÕ Sn
À» ±¸ÇϽÿÀ.
(´ä) (1/2)n(6n 2-3n-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) = 9 k2
-12 k +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)
- S =
(Á¦ 99Ç×) ÀÏ ¶§, S ¸¦ ±¸ÇϽÿÀ.¡¡
(´ä) 9

¢Ñ
- ¼ö¿ 1, 1+2, 1+2+3, 1+2+3+4, 1+2+3+4+5, ¡¦ ÀÇ Á¦10 Ç×±îÁöÀÇ ÇÕÀ» ±¸ÇϽÿÀ.
(´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 ^^
¢Ñ k(k+1) =
(1/3)n(n+1)(n+2), k(k+1)(k+2) = (1/4)n(n+1)(n+2)(n+3)
- S = 1¡¤10 + 3¡¤9 + 5¡¤8 +7¡¤7 + ¡¦ + 19¡¤1 ÀÏ ¶§, S ¸¦ ±¸ÇϽÿÀ.
(´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) = -2 k2
+23 k - 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
¸ñ·ÏÀ¸·Î
¡¡
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