第13题，数学，数列
\u6570\u5b66\u6570\u5217\u9898\u3001\u6025A(n+1)=2An+K
A(n)=2A(n-1)+K
A(n+1)-An=2[An-A(n-1)]
Bn=A(n+1)-An Bn-1= An-A(n-1)
Bn=2B(n-1)
{Bn}\u4e3a\u7b49\u6bd4\u6570\u5217
\u6c42\u8bc1 1/a(n-1)-1/an<1/n^2
a[n]=a[n-1]+a[n-1]^2/n^2>a[n-1]
n^2(a[n]-a[n-1])=a[n-1]^2
\u6240\u4ee51/a[n-1]-1/a[n]
=(a[n]-a[n-1])/a[n]a[n-1]
=a[n-1]/n^2[an]<1/n^2
\u5373\uff1a1/a[n-1]-1/a[n]<1/n^2
\u5f97\u8bc1\u3002(n+1)/(n+2)<an<n
\u6570\u5b66\u5f52\u7eb3\u6cd5\uff1a
n=1\u65f6\uff0c
a[1]\u4ee3\u5165\u5f97\uff1a(n+1)/(n+2)<a[1]<n\u6210\u7acb\u3002
\u8bben=k\u65f6\uff0c(k+1)/(k+2)<a[k]<k\u6210\u7acb\uff0c
\u5219\uff1an=k+1\u65f6\uff0c
a[k+1]=a[k]+a[k]^2/(k+1)^2
<k+k^2/(k+1)^2=k+1
>(k+1)/(k+2)+1/(k+2)^2
=(k^2+3k+3)/(k+2)^2
(k^2+3k+3)/(k+2)^2>(k+2)/(k+3)
\u5373\uff1a(k+2)^3<(k^2+3k+3) (k+3)
K^3+6k^2+12k+8<k^3+6k+12k+9\u2014\u2014\u6210\u7acb\uff01
\u6240\u4ee5(k+2)/(k+3)<a[k+1]<k+1
\u5373\u5bf9n=N\uff0c(n+1)/(n+2)<an<n\u90fd\u6210\u7acb\uff01
\u83aa\u624d\u521a\u4e0a\u521d\u4e00\uff0c\u4e0d\u61c2\u6ef4\u8bf4~\u8fd9\u662f\u767e\u5ea6\u4e0a\u641c\u6765\u7684\uff0c\u4e0d\u77e5\u9053\u5bf9\u4e0d\u5bf9\u3002\u3002
即Sn，S2n - Sn，S3n -S2n成等比数列。
于是(S2n -Sn)²=Sn·(S3n -S2n)
即12²=48(S3n -60)
解得 S3n=63
选D
设公比为q
Sn=a1*(1-q^n)/(1-q)=48 (1)
S(2n)=a1*[1-q^(2n)]/(1-q)=a1*(1+q^n)(1-q^n)/(1-q)=60 (2)
(2)/(1) 1+q^n=60/48=5/4 q^n=1/4
所以S(3n)=a1*[1-q^(3n)]/(1-q)
=[a1*(1-q^n)/(1-q)]*[1+q^n+q^(2n)]
=48*(1+1/4+1/4²)
=48+12+3
=63
希望能帮到你O(∩_∩)O
S3n+Sn=S2n*S2n
S3n=60*60/48=75,
选C
S2n/Sn=S3n/S2n
即60/48=S3n/60,
S3n=75
其实这个题目就是考查SN S2N-SN S3N-S2N 也是成等比的 一下子就能算出来了。 我这边下课了就不回答了 方法就是我上面所说的
绛旓細{lg a[n]}鏄瓑宸鏁板垪锛鎵浠 lg(a[n+1])-lg(a[n])=d 鎵浠 a[n+1]/a[n]=10^d 鏄笉鍙樼殑锛屾墍浠a[n]}鏄瓑姣旀暟鍒
绛旓細瑙:璁緓涓槦鍙傚姞姣旇禌,鍒 (x+1)x/2*2=90 瑙ｅ緱 x=9
绛旓細13.鐢眀n=|(an+2)/(an-1)|,鍙緱b(n+1)=|[a(n+1)+2]/[a(n+1)-1]|,鍐嶅皢a(n+1)=2/(an+1)甯﹀叆b(n+1)锛屽彲寰梑(n+1)=2bn,鎵浠n涓虹瓑姣鏁板垪锛鐢盿1=2,寰梑1=4,鎵浠n涓洪椤逛负4鍏瘮涓2鐨勭瓑姣旀暟鍒楋紝閫氶」涓篵n=4*2^(n-1)=2^(n+1)14.鏄撶煡绛夊樊鏁板垪鍓峮椤瑰拰Sn=(d/2...
绛旓細绠鍗曞垎鏋愪竴涓嬶紝绛旀濡傚浘鎵绀
绛旓細瑙ｏ細鈭礢n =7n+n(n1)d/2 锛屽綋涓斾粎褰搉=8鏃禨n鍙栧緱鏈澶у硷紝鈭 S7锛淪8 S9锛淪8 锛屽嵆 49+21d锛56+28d 63+36d锛56+28d 锛岃В寰楋細d锛1 d锛7/8 缁间笂锛歞鐨勫彇鍊艰寖鍥翠负锛-1锛-7/8锛夛紟
绛旓細1.棣栧厛鑰冭檻鍏瘮q=1鐨勬儏鍐碉細鍒檃[n+1]=a[1]锛孲[n]=n*a[1]锛宯瓒嬭繎浜庢棤绌锋椂锛屼袱鑰呬箣姣斾负n鍒嗕箣1锛屽嵆0锛屾弧瓒抽鎰忋2.褰撳叕姣攓涓嶇瓑浜1鐨勬椂鍊欙細鍒檃[n+1]=a[1]*q^n锛孲[n]=a[1]*(1-q^n)鍐嶉櫎浠ワ紙1-q锛夛紙娉細涓轰簡闃叉鐞嗚В闂棰橈紝鎴戝啀鐢ㄦ眽瀛楀啓涓閬嶏細a[n+1]绛変簬a[1]涔樹互鍏瘮q...
绛旓細褰 x+y=1 鏃讹紝f(x)+f(y) = (2^x+2^y+2鈭2) / [2^(x+y)+鈭2*2^x+鈭2*2^y+2] = 1/鈭2 锛屽洜姝 f(-5)+f(-4)+...+f(0)+...+f(6)= [f(-5)+f(6)]+[f(-4)+f(5)]+...+[f(0)+f(1)]= 6/鈭2 = 3鈭2 銆
绛旓細鍥犳棰樼洰璇村弬鑰冪瓑宸鏁板垪鐨勬柟娉曪紝鑰岄涓-11锛屽熬涓12锛-11+12=1锛屽洜姝よ寰嬫х殑绠楃畻 f(a)+f(1-a)鐨勫硷紝鐪嬫槸涓嶆槸甯告暟锛岃繕鏄湁浠涔堣寰 褰撶劧鏈夋椂鍊欐病浠涔堬紝浣嗘槸杩欎笉浠ｈ〃鏂规硶涓嶈兘鐢紝濡傛灉绠楀嚭鏉ョ殑涓嶆槸甯告暟锛屽彲浠ユ崲涓濊矾鑰冭檻涓涓 灏辨嬁鏈鏉ヨ锛屽鏋渇(a)+f(1-a)涓嶆槸甯告暟锛屼綘鍙冭檻f(a)+f...
绛旓細x)g'(x)<0锛屽嵆F'(x)<0锛屾墍浠0<a<1銆傜劧鍚庡浣曟眰a绛旀缁欑殑寰堣缁嗭紝搴旇涓嶇敤鍐嶈В閲婁簡鍚с鏁板垪鐨勫墠k椤瑰拰涓篠k锛屽綋k鍦1鍒10杩10涓暟涔嬮棿鍙栧兼椂锛屾湁鍑犱釜鍙互浣縎k>15/16锛屾眰鍑虹鍚圫k>15/16鐨刱鐨勪釜鏁帮紝鐒跺悗闄や互10灏辨槸鎵姹傛鐜囧晩銆傜瓟妗堝叾瀹炵粰鐨勫緢璇︾粏浜嗭紝鍙槸涓嶇煡閬撲綘鍝噷涓嶆噦锛屽彲浠ヨ拷闂 ...
绛旓細杈归暱涓2鐨勬槸n鈥1涓 2n+2涓烘墍鏈夐暱鏂瑰舰鐨勪釜鏁 闀挎柟褰㈢殑涓暟n+(n+1锛+鈥︹+1=2n+2 鏁板浘褰㈢殑瑙勫緥 杩欎釜灏忓濂ユ暟鐨勬暟鍥惧舰涓撻鏈夛紝姹傚緱杩囩▼闇瑕佺敤绛夊樊鏁板垪 瀛╁瓙锛屼綘閲囩撼鐨勬槸閿欒绛旀锛屼綘鏄庡ぉ鍙鎵ｅ垎浜嗗摝
