错误！未找到引用源。解:(1)设A0(?p， y0),由于青蛙依次向右向上跳动, 2
y0),A2(， ?y0),由抛物线定义知:S2?3p所以A1(，(2) 依题意,x2n?1?n??p2p2x2n?1， x2n?x2n?1， y2n?y2n?1?x2n?1(n?N*)
limSn?|A0A1|?|A1A2|?|A2A3|?|A3A4|???|A2n?2A2n?1|?|A2n?1A2n|??
?(x1?x0)?(y2?y1)?(x3?x2)?(y4?y3)?(x5?x4)???(x2n?1?x2n)?(y2n?y2n?1)???2(x1?x0)?2(x3?x2)?2(x5?x4)???2(x2n?1?x2n)??
， 1) 随着n的增大,点An无限接近点(1
横向路程之和无限接近1?所以 limSn=
n???1111?,纵向路程之和无限接近1?? 222211??1 22(注:只要能说明横纵坐标的变化趋势,用文字表达也行)
(3)设点A2k(x2k， y2k)， A2k?1(x2k?1， y2k?1),由题意,An的坐标满足如下递推关系:
1x0?， y0?1,且y2k?1?y2k(k?0， 1，，， 2 3 ?)， x2k?1?x2k?2(k?0， 1，，， 2 3 ?)
2其中y2k?1?x2k?1， y2k?2x2k,∴x2k?1?x2k?2?2x2k, (方法一)∴{x2k}是以x0?11k为首项,2为公比的等比数列,∴x2k??2,y2k?2k 22n1n即当n为偶数时,xn??22,yn?22
2n?12n?12又x2k?1?x2k?2?2,y2k?1?x2k?1?2,∴当n为奇数时,xn?2于是,当n为偶数时,
kk， yn?2
|A0A1|?|A1A2|?|A2A3|?|A3A4|???|A2k?2A2k?1|?|A2k?1A2k|
?(x1?x0)?(y2?y1)?(x3?x2)?(y4?y3)?(x5?x4)???(x2k?1?x2k?2)?(y2k?y2k?1)
?(x1?x0)?(y2?y0)?(x3?x1)?(y4?y2)?(x5?x3)???(x2k?1?x2k?3)?(y2k?y2k?2)33?(x2k?y2k)?(x0?y0)??2k?
22当n为奇数时,
|A0A1|?|A1A2|?|A2A3|?|A3A4|???|A2k?2A2k?1|?|A2kA2k?1|
?(x1?x0)?(y2?y1)?(x3?x2)?(y4?y3)?(x5?x4)???(y2k?y2k?1)?(x2k?1?x2k)
?(x1?x0)?(y2?y0)?(x3?x1)?(y4?y2)?(x5?x3)???(y2k?1?y2k)?(x2k?1?x2k?1)?(x2k?1?y2k?1)?(x0?y0)?2?2k?3 2?1?n232? n为奇数?2∴Sn?? n3?(22?1) n为偶数?211kk(方法二)∴{x2k}是以x0?为首项,2为公比的等差数列,∴x2k??2,y2k?2
22
又x2k?1?x2k?2?2k,y2k?1?x2k?1?2k ∴x2k?1?x2k?2?k1k1k?2??2,y2k?2?y2k?1?2k?1?2k?2k 22于是,当n为偶数时,
|A0A1|?|A1A2|?|A2A3|?|A3A4|???|A2k?2A2k?1|?|A2k?1A2k|
?(x1?x0)?(y2?y1)?(x3?x2)?(y4?y3)?(x5?x4)???(x2k?1?x2k?2)?(y2k?y2k?1)
1133?(?1?2????2k?1)?(1?2???2k?1)??2k? 2222当n为奇数时,
|A0A1|?|A1A2|?|A2A3|?|A3A4|???|A2k?2A2k?1|?|A2kA2k?1|
?(x1?x0)?(y2?y1)?(x3?x2)?(y4?y3)?(x5?x4)???(y2k?y2k?1)?(x2k?1?x2k)
113?(?1?2????2k)?(1?2???2k?1)?2?2k? 222?1?n232? n为奇数?2∴Sn?? . n3?(22?1) n为偶数?2(注:本小题若没有写出递推关系,直接归纳得到正确结论而没有证明,扣4分)
错误！未找到引用源。解:(1)?an?是等差数列,∴
2222013?(a?b)?2013,即a?b?2
2所以c?a?b???2,c的最小值为2;
(2)设a,b,c的公差为d(d?Z),则a2?(a?d)2?(a?2d)2?a?3d 设
三
角
形
的
三
边
长
为
3d,4d,5d,面积,
Sd?13?d4?2d26?d(d?,)ZSn?6n2T2n??S1?S2?S3???S2n?6[?12?22?32?42???(2n)2] ?6(1?2?3?4????2n)?12n2?6n
1n?2n, 2n(n?1)1n???2?2n?(n2?n)?n2?n, 当n?5时,2?1?n?22由T2n?6?2n?1得n?2
经检验当n?2,3,4时,n?211n?2n,当n?1时,n2?n?2n 22综上所述,满足不等式T2n?6?2n?1的所有n的值为2、3、4 (3)证明:因为a,b,c成等比数列,b?ac.
2由于a,b,c为直角三角形的三边长,知a?ac?c,
nn22c1?5, ?a2nn?1?5??1?5?c??a???????, 又5Xn??,得5X?n??????(n?N)?????a??c??2??2??1?5??1?5??1?5??????于是5Xn?5Xn?1???2???2???2????????1?5?????2???n?2nnn?1?1?5?????2???n?1
?1?5?????2???n?2?5Xn?2
?Xn+Xn?1?Xn?2,则有?故数列
?Xn??2+Xn?1??2?Xn?2?.
2?Xn中的任意连续三项为边长均可以构成直角三角形
?1122???????????5?11?5?,5?5?11?5?? 因为 X1?5??=1X?????????????2??2??2??2??=15??25????????????????X3?X1?X2?2?N?,
由Xn?Xn?1?Xn?2,同理可得Xn?N?,Xn?1?N??Xn?2?N?, 故对于任意的n?N都有Xn是正整数
错误！未找到引用源。 [解] (1)如图,由?OQ1P1是边长为a1的等边三角形,得点P1的坐标
?23a12a1a13a1a13a12?,得a1? ),又P1(,)在抛物线y?x上,所以为(,3422222同理P2(?234a23a2,?)在抛物线y2?x上,得a2?
3221:点Qn?1的坐标为(a1?a2?a3?????an?1,0),即点
(2)如图,法
(Sn?1,点0)Q(与原点重合，0S0,所以直线=0)Qn?1Pn的方程为y?3(x?Sn?1)或
y??3(x?S),因此,点P??y2?xn?1n的坐标满足? ??y?3(x?Sn?1)消去x得3y2?y?3Sn?1?0 , 所以y?1?1?12Sn?123 又y?a3n?sin60??2an,故3an?1?1?12Sn?1 从而3a2n?2an?4Sn?1 ① 由①有3a2n?1?2an?1?4Sn ②
②-①得3(a22n?1?an)?2(an?1?an)?4an
即(an?1?an)(3an?1?3an?2)?0,又an?0,于是a2n?1?an?3 所以{a2n}是以
3为首项、23为公差的等差数,aa2n?1?(n?1)d?3n S(a1?an)n1n?2?3n(n?1)
G3232Gn3n23n?4an?9n,lim 理2分 n??S?lim?nn??3n(n?1)3法
2:
点
Qn?1的坐标为
(a1?a2?a3?n??,?a,0(Sn?1,点0与原点重合，Q)0(S, 0=0)所以直线Qn?1Pn的方程为y?3(x?Sn?1)或y??3(x?Sn?1)
因此,点P??y2?xn(x,y)的坐标满足?消去y得3(x?S??y?3(x?Sn?1)2?x, n?1)又x?Sana2a2n?1?2,所以3(n2)?Sn?1?n2,从而3an?2an?4Sn?1 ①
以下各步同法1
法3:
点Qn?1的坐标为(a1?a2?a3?????an?1,0),
即点(Snn?1,0)(点Q0与原点重合，S0=0),所以Pn(Sn?1?a2,3an2), 又Pan3an32n(S2,2)在抛物线y2?x上,得?Sann?1?4ann?1?2
即)?点
?
2即3an?2an?4Sn?1
以下各步同法1
ba(3)(理)因为n?1?bn2(n?1)3a2n3?a,
232323所以数列{bn}是正项等比数列,且公比q0?a?1,首项b1?a?q0,
qrsb1(1?q0p)b1(1?q0)b1(1?q0)b1(1?q0)则Tp?,Tq?,Tr?,Ts?
1?q01?q01?q01?q0b12sqrq?r(注意q0p?s?q0) Tp?Ts?Tq?Tr=??(1?q0p)(1?q0)?(1?q0)(1?q0)?2??(1?q0)b12qrps? ??(q?q)?(q?q)?00002??(1?q0)qrsqsr而(q0?q0)?(q0p?q0)?(q0?q0p)?(q0?q0)
q?prs?rq?pr?q0p(q0?1)?q0(q0?1)?(q0?1)(q0p?q0)(注意q?p?s?r) q?pr?pq?pr?p?(q0?1)q0p(1?q0)??q0p(q0?1)(q0?1)
23因为a?0且a?1,所以q0?a?0且q0?1
q?pr?p又q?p,r?p均为正整数,所以(q0?1)与(q0?1)同号, pq?pr?p故?q0(q0?1)(q0?1)?0,所以,Tp?Ts?Tq?Tr
(第(3)问只写出正确结论的,给1分)