两道等比高一数学题,高手来来来。
\u9ad8\u4e00\u6570\u5b66\u7b49\u6bd4\u6570\u5217\u9898\u76ee1.
\u2235sn=a1(1-q^n)/(1-q)=(a1-q\u00d7an)/(1-q)
\u2234189=(3-96q)/(1-q)
\u2234q=2
\u223596\uff1d3\u00d72^(n-1)
\u2234n=6
2.
\u2235s2\uff1da1\uff081\uff0dq²\uff09\uff0f\uff081\uff0dq\uff09\uff1d7
s4\uff1da1\uff081\uff0dq^4\uff09\uff0f\uff081\uff0dq\uff09\uff1da1\uff081\uff0bq²\uff09\uff081\uff0dq²\uff09\uff0f\uff081\uff0dq\uff09
s6\uff1da1\uff081\uff0dq6\uff09\uff0f\uff081\uff0dq\uff09\uff1da1\uff081\uff0dq²\uff09\uff081\uff0bq²\uff0bq^4\uff09\uff0f\uff081\uff0dq\uff09\uff1d91
\u2234s6\u00f7s2\uff1d1\uff0bq²\uff0bq^4\uff1d13
\u2234q²\uff1d3
\u2235s4\u00f7s2\uff1d1\uff0bq²\uff1d4
\u2234s4\uff1d4s2\uff1d28
3.
\u2235an\uff1d2^n-1
\u2234a1\uff0ba2\uff0b\u2026\u2026\uff0ban\uff1d\uff082\uff0d1\uff09\uff0b\uff082²\uff0d1\uff09\uff0b\u2026\u2026\uff0b\uff082^n\uff0d1\uff09
\uff1d\uff082\uff0b2²\uff0b\u2026\u2026\uff0b2^n\uff09\uff0dn
\uff1d2^(n\uff0b1\uff09\uff0d2\uff0dn
4.
\u505a\u6cd5\u7c7b\u4f3c\u4e8e2,\u8be6\u7ec6\u8fc7\u7a0b\u6211\u5c31\u4e0d\u5199\u51fa\u6765\u4e86\u3002
\u2235s2n\u00f7sn\uff1d1\uff0bq^n\uff1d5
\u2234q^n\uff1d4
\u2235s3n\u00f7sn\uff1d1\uff0bq^n\uff0bq^2n\uff1d21
\u2234s3n\uff1d21sn\uff1d42
\u5217\u8868
x 1 2 3....10
y 9 15 32..36
\u5b9a\u4e49\u57df\u30101,10\u3011
\u503c\u57df{9,15,32\u3002\u3002\u300236\uff08\u5373\u4e3ay\u7684\u53d6\u503c\uff09
S1=a1
S2=a1+a2=2a1+d
S4=a1+a2+a3+a4=4a1+6d
因为成等比数列 ，所以S2的平方=S1*S4
（2a1+d)的平方=a1(4a1+6d)
因为d不得0
解得d=2a1
所以S2=4a1
q=S2/S1=4
因为S2=6
所以4a1=6
a1=3/2
d=2a1=3
an=a1+(n-1)d=3/2+(n-1)*3=3n-1.5
第二题：
A2=A1+D，A4=A1+3D，（其中D为公差），∵A1,A2,A4这三项构成等比数列，∴（A1+D）的平方=A1×（A1+3D），有D的平方+2×A1×D=3×A1×D，则，A1=DA1=D，A2=2D，A4=4D，易知，公比Q为2。
s1=a1
s2=a1+a2
s4=a1+a2+a3+a4
s2^2=s1*s4
a1^2(1+q)^2=a1^2(1+q+q^2+q^3)
(1+q)^2=[(1+q)(1+q^2)]
(1+q)(1+q+1+q^2)=0
q=-1或q1=...
1，公比为2，S1=a,S2=2a,S4=4a
a=3，an=3n
2，q=2
2
3*6(n-1)
2
设等差数列第一项为a1,公差为d
S1=a1
S2=2a1+d
S4=4a1+6d
S1,S2,S4成等比数列，即
(2a1+d)^2=a1(4a1+6d)
解出d=2a1
q=S2/S1=(2a1+d)/a1=4
若S2=6,则a1=1.5,d=3
an=1.5+3(n-1)=3n-1.5
2、
设公差为d
a2=a1+d
a4=a1+3d
a1，a2,a4成等比数列，即
(a1+d)^2=a1(a1+3d)
解出d=a1
q=a2/a1=(a1+d)/a1=2
S1=a1
S2=a1+a2=a1+a1+d=2a1+d
S4=a1+a2+a3+a4=4a1+(1+2+3)d=4a1+6d
S2^2=S1×s4
(2a1+d)^2=a1×(4a1+6d)
a1×q=2a1+d
a1×q^2=4a1+6d
[(2a1+d)÷a1]^2=(4a1+6d)÷a1
4a1^2+4a1d+d^2=4a1^2+6a1d
2a1d-d^2=0 即2a1=d
S2=2a1+d=2d=6 d=3 a1=3/2
等差数列的通项公式 3n-3/2
q=(2a1+d)/a1=4
2 a2^2=a1×a4
(a1+d)^2=a1×(a1+3d) a1=d
q=a2/a1=(a1+d)/a1=(2a1)/a1=2
绛旓細an=a1+(n-1)d=3/2+(n-1)*3=3n-1.5 绗浜岄锛欰2=A1+D锛孉4=A1+3D锛岋紙鍏朵腑D涓哄叕宸級锛屸埖A1,A2,A4杩欎笁椤规瀯鎴绛夋瘮鏁板垪锛屸埓锛圓1+D锛夌殑骞虫柟=A1脳锛圓1+3D锛夛紝鏈塂鐨勫钩鏂+2脳A1脳D=3脳A1脳D锛屽垯锛孉1=DA1=D锛孉2=2D锛孉4=4D锛屾槗鐭ワ紝鍏瘮Q涓2銆
绛旓細an=n(n+1)(2n-1)=2n³+n²-n Sn=a1+a2+...+an =2(1³+2³+...+n³)+(1²+2²+...+n²)-(1+2+...+n)=2[n(n+1)/2]²+n(n+1)(2n+1)/6-n(n+1)/2 =[n(n+1)/6][3n(n+1)+(2n+1) -3]=[n(n+1...
绛旓細{an}涓绛夋瘮鏁板垪锛宐n=na1+(n-1)a2+鈹勨攧+2a(n-1)+an锛宐1=m锛宐2=3m/2锛涳紙1锛塨1=a1=m锛宐2=2a1+a2=3m/2锛a2=-m/2锛宷=-1/2锛屾暟鍒梴an}棣栭」涓簃锛屽叕姣攓=-1/2锛涳紙2锛塵=1锛宎n=(-1/2)^(n-1)锛宐n=n(-1/2)º+(n-1)(-1/2)+鈹勨攧+2(-1/2)^(n-2)+(-1/2...
绛旓細2銆傚厛鍊熶粬浠4鍙嫻鏋滐紝閭ｄ箞绗竴鍙尨瀛愬皢鑻规灉鎭板垎浣滀簲浠芥嬁璧颁竴鍫嗭紙鍏跺疄杩欎竴鍫嗕篃灏辨槸鍏堝墠浠栨嬁鐨勪竴鍫嗗姞涓涓級锛屽墿涓嬮偅鍥涘爢鍚勫彇鍑轰竴涓浜鍙尨瀛愭潵鍒嗚嫻鏋滐紝灏嗗洓鍙嫻鏋滃張鏀捐繘鍘伙紝鍥犱负鍘熸湰澶氫竴涓嫻鏋滐紝灏卞彲浠ュ钩鍒嗘垚鈥︹︾瓟妗堝氨鏄5*5*5*5*5-4=3121 杩欎釜鏄 鏈灏忓 绗竴鍙尨瀛愬悆鎺変竴涓墿3120 鎷胯蛋...
绛旓細瑙9鐢遍鐭 am锛宎锛坢+5锛夛紝a锛坢+10锛夛紝a锛坢+15锛夋垚绛夋瘮鏁板垪 鐢盿m=3 a锛坢+5锛=24 鐭锛坢+10锛=192 a锛坢+15锛=1536 10鐢盿1=2锛a3=a1+2d=2+2d锛宎11=a1+10d=2+10d 鍙堢敱a1a11=(a3锛塣2 寰2锛2+10d锛=锛2+2d锛塣2 鍗筹紙1+5d锛=锛坉+1锛塣2 鍗砫^2-3d=0 瑙ｅ緱d=3鎴朼...
绛旓細S2n = a1 * (1-q^2n) / (1-q)S2n / Sn = 1+q^n = 6560/80 = 82 鎵浠鐨刵娆℃柟鏄81锛岃繖鏍风殑鏉′欢涓嬫湁鍙兘q=9,n=2鎴栬卶=3,n=4 浣嗘槸鍓嶈呬娇寰楀墠2椤瑰垎鍒槸8鍜72锛屼笌鏈澶х殑椤规槸54鐭涚浘锛岃垗鍘汇傛墍浠=3, n=4锛孉n = A1 * 3^(n-1)鎵浠4 = 27 * A1 = 54锛孉1=2 鍓4椤...
绛旓細1.鈭绛夋瘮 鈭碼2a4=a3^2 a4a6=a5^2 鍘熷紡=a3^2+a5^2+2a3a5=25 鍙坅n>0 鈭碼3+a5=5 2.a6=(q^3)a3,a7=(q^3)a4,a8=(q^3)a5 a6a7a8=(q^9)a3a4a5 q^9=8 q9a10a11=(q^9)a6a7a8 =192
绛旓細1銆乹³=a4梅a1=-64=锛-4锛³q=-4 a2=a1q=4锛宎3=a2q=-16 S4=a1+a2+a3+a4=-1+4-16+64=51 2銆乤1+a2=a1锛1+q锛=S3-a3=3 鑰宎3=a1q²=3/2 鈭磓²/锛1+q锛=1/2 2q²-q-1=0 锛2q+1锛夛紙q-1锛=0 q=-1/2鎴杚=1 鑻=-1/2锛姝ゆ椂a1...
绛旓細绗竴閬撻 0锛绛夋瘮鏁板垪 a aq aq² 鍥犳鏂圭▼涓篴x²+aqx+aq²=0锛屾眰瑙ｇ殑涓暟灏辨槸闂 b²-4ac鏄>0,<0,杩樻槸=0锛屽洜姝²-4ac=锛坅q锛²-4a锛坅q锛²=鈥3a²q²<0,--->鏃犺В锛屽嵆涓巟杞存病鏈変氦鐐癸級绗浜岄亾棰 84(91-7=84)绗笁閬...
绛旓細鐢卞叕寮忥細a1+a1*q+a1*q*q+a1*q*q*q+...+a1*q^n=a1(1-q^n)/(1-q)寰楋細1+a+a^2+...+a^10=1*(1-a^n)/(1-a)鎴栬(a^n-1)/(a-1)锛岄兘涓鏍枫傚笇鏈涘浣犳湁鎵甯姪銆
