Sn为等比数列前n项和,为什么不能说Sn,S2n-Sn,S3n-S2n为等比数列 公比不为-1的等比数列{an}的前n项和为Sn，则Sn,S2...
\u7b49\u6bd4\u6570\u5217an\u7684\u524dn\u9879\u548c\u4e3asn(sn\u22600),\u5219sn,s2n-sn,s3n-s2n\u6210\u7b49\u6bd4\u6570\u5217,\u516c\u6bd4\u4e3a\uff1fSn=a1+a2+a3+......+an S2n-Sn=a(n+1)+a(n+2)+a(n+3)+......+a2n =a1*q^n+a2*q^n+a3*q^n+......+an*q^n =(q^n)*(a1+a2+a3+.....+an) =Sn*q^nS3n-S2n =a(2n+1)+a(2n+2)+a(2n+3)+.....+a3n =a1 *q^2n+a2*q^2n+a3*q^2n+.....+an*q^2n =(q^2n)\uff08a1+a2+a3+....+an\uff09 =Sn*q^2nSn*(S3n-S2n)=(Sn^2)*(q^2n)(S2n-Sn\uff09²=\uff08 Sn^2)*(q^2n) Sn*S3n-S2n=\uff08S2n-Sn\uff09²\uff0c\u6240\u4ee5Sn,S2n-Sn,S3n-S2n\u6210\u7b49\u6bd4\u6570\u5217\uff0c\u516c\u6bd4\u4e3aq^n.
\u7528\u7b49\u6bd4\u6570\u5217\u7684\u6c42\u548c\u516c\u5f0f\u7b80\u5355\u8ba1\u7b97\u4e0b\u5c31\u80fd\u5f97\u5230\u4e86\u3002
因为等比数列的任何一项都不能为零，但Sn,S2n-Sn,S3n-S2n这三项中不能保证都不为零，所以不能说Sn,S2n-Sn,S3n-S2n为等比数列因为sn=(a1-an*q)/(1-q)而S2n-Sn=(-a2n*q+an*q)/(1-q)=anq(-q(n次方)+1)/(1-q)即他们不符合等比数列的定义，当然不是等比数列咯！
sn
=
a1
*
(q^n
-1)/(q-1)
s2n
=
a1
*
(q^2n
-1)/(q-1)
s<(k-1)n>
=
a1
*
[q^(k-1)n
-1]/(q-1)
skn
=
a1
*
[q^(kn)
-1]/(q-1)
s<kn>
-
s<(k-1)n>
=[a1/(q-1)]*[q^(kn)
-
q^(k-1)n]
=
[a1/(q-1)]
*
q^[(k-1)n]
*
(q^n
-1)
=
[a1
*
(q^n
-1)/(q-1)]
*
q^[(k-1)n]
=
sn
*
(q^n)^(k-1)
从上面表达式已经可以直接看出,
它恰好为等比数列的通项公式
首项为
sn,
公比为
q^n
因此
sn,s2n-sn,s3n-s2n......成等比数列
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绛旓細Sn=[a1*(1-q^n)]/(1-q)涓虹瓑姣旀暟鍒鑰岃繖閲宯涓烘湭鐭ユ暟鍙互鍐欐垚F锛坣锛=[a1*(1-q^n)]/(1-q)褰搎=1鏃朵负甯告暟鍒椾篃灏辨槸n涓猘1鐩稿姞涓簄*a1銆傚鏋滀竴涓暟鍒椾粠绗2椤硅捣锛屾瘡涓椤逛笌瀹冪殑鍓嶄竴椤圭殑姣旂瓑浜庡悓涓涓父鏁帮紝杩欎釜鏁板垪灏卞彨鍋氱瓑姣旀暟鍒椼傝繖涓父鏁板彨鍋氱瓑姣旀暟鍒楃殑鍏瘮锛屽叕姣旈氬父鐢ㄥ瓧姣峲琛ㄧず(q鈮0)銆
绛旓細瑙ｏ細绛夋瘮鏁板垪{an}鐨鍓峮椤瑰拰涓Sn锛宸茬煡S3=4锛孲6=12锛屽垯鐢辩瓑姣旀暟鍒楃殑鎬ц川鍙緱锛歋3 銆丼6-S3銆丼9-S6浠嶆垚绛夋瘮鏁板垪锛鍗4锛12-4锛孲9-12鎴愮瓑姣旀暟鍒楋紝鏁呮湁 锛12-4锛夛紙12-4锛=4锛圫9-12 锛夆埓S9=28 甯屾湜鎴戠殑鍥炵瓟瀵规偍鏈夊府鍔╋紝鏈夐棶棰樺彲浠ヨ拷闂傛弧鎰忚鍙婃椂閲囩撼锛岃阿璋紒
绛旓細瀵逛簬绛夋瘮鏁板垪锛鑻ラ椤规槸 a锛屽叕姣旀槸 r锛屽垯绗 n 椤瑰彲琛ㄧず涓 a * r^(n-1)銆傞鍏堬紝鎴戜滑璁＄畻绛夋瘮鏁板垪鐨鍓 n 椤瑰拰 Sn锛歋n = a + a * r + a * r^2 + ... + a * r^(n-1)鐒跺悗锛岃绠楃瓑姣旀暟鍒楃殑鍓 2n 椤瑰拰 S2n 鍜屽墠 3n 椤瑰拰 S3n锛歋2n = a + a * r + a * r^2 + ....
绛旓細Sn,S2n-Sn,S3n-S2n鎴绛夋瘮鏁板垪,鍏瘮涓簈^n.璇佹槑:鍏堣瘉鏄庝竴涓洿涓鑸殑閫氶」鍏紡.鍦ㄧ瓑姣旀暟鍒椾腑,an=a1q^(n-1)am=a1q^(m-1)涓ゅ紡鐩搁櫎寰梐n/am=q^(n-m),鈭碼n=amq^(n-m).S2n=a1+a2+...+an+a(n+1)+a(n+2)+...+a2n =Sn+(a1q^n+a2q^n+...+anq^n)=Sn+(a1+a2+...+an...
绛旓細绛夋瘮鏁板垪姹傚拰鍏紡锛氭眰鍜屽叕寮忕敤鏂囧瓧鏉ユ弿杩板氨鏄細Sn=棣栭」锛1-鍏瘮鐨刵娆℃柟锛/1-鍏瘮锛堝叕姣斺墵1锛夊鏋滃叕姣攓=1锛屽垯绛夋瘮鏁板垪涓瘡椤归兘鐩哥瓑锛屽叾閫氶」鍏紡涓 锛屼换鎰忎袱椤 锛 鐨勫叧绯讳负 锛涘湪杩愮敤绛夋瘮鏁板垪鐨鍓峮椤瑰拰鏃讹紝涓瀹氳娉ㄦ剰璁ㄨ鍏瘮q鏄惁涓1銆
绛旓細绛夋瘮鏁板垪鐨鍓峮椤瑰拰鍏锋湁褰㈠紡sn=a1*(1-qn)/(1-q)锛岃岀敱鍙宠竟棰樻剰鍙互鐭ラ亾鏁板垪{sn}鐨勯氶」鍏紡涓2*2锛坣-1锛=2鐨刵娆″箓锛岃瀵熶竴涓嬩袱涓紡瀛愬氨鍙互鍙戠幇浠栦滑鏈夌浉鍚岀殑鍦版柟锛屾墍浠ユ暟鍒梴an}鑳芥垚绛夋瘮鏁板垪 渚嬪鍙朼1=1锛宷=2灏辨槸浜
