(1)
è¯æï¼ç±å¼a(n+1)=2an+2^nå·¦å³ä¸¤è¾¹åæ¶é¤ä»¥2^n,å¯å¾:a(n+1)/2^n=a(n)/2^(n-1)+1 â
ãã åé¢ä¸ä»¤bn=an/2^(n-1),åb(n+1)=a(n+1)/2^n,å°b(n)åb(n+1)带å
¥â å¼å¾b(n+1)=b(n)+1
å³ b(n+1)-b(n)=1 æ以æ°å{bn}为çå·®æ°å
(2)
解ï¼a(n+1)/2^n=2an/2^n+1
â´a(n+1)/2^n=an/2^(n-1)+1
â´a(n+1)/2^n-an/2^(n-1)=1
â´æ°æ°å{an/2^(n-1)}å°±æäºä¸ä¸ªä»¥a1/2^0=1为é¦é¡¹ 1为å
¬å·®ççå·®æ°å
â´an=nÃ2^(n-1)
â´Sn=a1+a2+...+an
=1.2^0+2.2^1+...+n.2^(n-1) (1)
â´2Sn= 1.2^1+2.2^2+...+ï¼n-1).2^(n-1)+n.2^n ï¼2ï¼
ï¼1ï¼-ï¼2ï¼å¾ï¼-Sn=2^0+2^1+2^2+...+2^(n-1)-n.2^n
=1.(1-2^n)/(1-2)-n.2^n
=2^n-n.2^n-1
â´Sn=(n-1).2^n+1
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