已知等差数列{an}中a1=2,a5=10.求数列{an}的通项公式an拜托各位大神
已知等差数列{an}中a1=2,a5=10,有:a5=a1+4d=2+4d=10 d=2 an=2+2(n-1)=2n绛旓細绛夊樊鏁板垪閫氶」鍏紡涓猴細an=a1+锛坣-1锛壝梔 绛夊樊鏁板垪姹傚拰鍏紡涓猴細Sn=锛坅1+an锛壝梟梅2 浠ヤ笂鍏紡涓璬涓哄叕宸紝n涓洪」鏁般傝В锛a1=2锛宎3=2 a3=a1+锛坣-1锛壝梔=2+锛3-1锛壝梔=2 瑙e緱d=0 a8=a1+锛坣-1锛壝梔=2+锛8-1锛壝0=2 鎵浠8=锛坅1+a8锛壝8梅2 =锛2+2锛壝8梅2 =16 ...
绛旓細•闂锛绛夊樊鏁板垪{an}涓璦1=2锛宎5=8锛屾眰an鍜宻n銆•瑙g瓟锛氭牴鎹鎰忥紝瀵逛簬绛夊樊鏁板垪锛璁剧瓑宸涓篸锛屽垯鏈夛細 a5=a1+(5-1)*d=2+4*d=8,鎵浠4d=8-2=6锛屾眰鍑篸=3/2,姹傚叕宸甦 杩涗竴姝ョ敱绛夊樊鏁板垪鎬ц川鍙煡锛氳绛夊樊鏁板垪鐨勯氶」an=a1+(n-1)*d=2+(n-1)*3/2=(3n+1)/2銆傝繘涓姝ュ彲...
绛旓細鍒嗘瀽锛氬厛鏍规嵁a1=2锛宎2+a3=13姹傚緱d鍜宎5锛岃繘鑰屾牴鎹瓑宸腑椤圭殑鎬ц川鐭4+a5+a6=3a5姹傚緱绛旀锛庤В绛旓細瑙o細鍦绛夊樊鏁板垪{an}涓锛宸茬煡a1=2锛宎2+a3=13锛屽緱d=3锛宎5=14锛屸埓a4+a5+a6=3a5=42锛庢晠閫塀 鐐硅瘎锛氭湰棰樹富瑕佽冩煡浜嗙瓑宸暟鍒楃殑鎬ц川锛庡睘鍩虹棰橈紟
绛旓細瑙o細宸茬煡绛夊樊鏁板垪{an}涓紝a1=2锛宎1+a2=a3 璁惧叕宸负 d 鍒 a2=a1+d=2+d锛宎3=a1+2d=2+2d 2+2d=2+2+d 寰 d=2 鎵浠 an=a1+(n-1)d=2+2(n-1)=2n
绛旓細瑙o細鐢遍鍙煡鍏樊d=A2-A1=6锛涙墍浠3=A2+d=14锛汚4=A3+d=20锛涙墍浠4=A1+A2+A3+A4=2+8+14+20=44
绛旓細d=(a3-a1)/2=(10-2)/2=4 鎵浠 an=a1+(n-1)d=2+4(n-1)=4n-2 a8=4脳8-2=30 s8=(a1+a8)脳8梅2=锛2+30锛壝4=128
绛旓細鍦绛夊樊鏁板垪{ an}涓 宸茬煡a1=2,a2=4, 閭d箞a5绛変簬10.
绛旓細鍦鏁板垪{an}涓锛a1=1锛屽綋n澶т簬绛変簬2鏃讹紝鍏跺墠n椤瑰拰sn婊¤冻an=锕2SnS锛坣-1锛 an=锕2SnS锛坣-1锛 an=Sn-S锛坣-1锛 Sn-S锛坣-1锛=锕2SnS锛坣-1锛 涓よ竟鍚岄櫎浠nS锛坣-1锛 1/S(n-1)-Sn=锕2 1/Sn-1/S(n-1)=2 鏁板垪{1/Sn}鏄绛夊樊鏁板垪 1/S1=1/a1=1 1/Sn=1+2(n...
绛旓細S7=35,n=7,a1=2 S7=7锕2+7锕6/2锕=35 2+3锕=5 d=1
绛旓細鍦绛夊樊鏁板垪{ an}涓 鈭a1=2,a2=4,鈭碼2=a1+1脳d 鈭磀=2 鈭碼5=a1+4d=2+4脳2=10