若数列{bn}满足b1=1,b2=2,bn+2=3bn+1-2bn,求{bn}的通项公式.
问题描述:
若数列{bn}满足b1=1,b2=2,bn+2=3bn+1-2bn,求{bn}的通项公式.
答
bn=2^(n-1)b(n+2)-b(n+1)=2(b(n+1)-bn)依次累乘得(b(n+2)-b(n+1))/(b2-b1)=2^n即b(n+2)-b(n+1)=2^n.b2-b1=2^0累加bn-b1=2^(n-2)+...+2^0bn=2^(n-2)+...+2^0+2^02bn=2^(n-1)+...+2^1+2^1相减bn=2^(n-1)+2^1-2^0-2^0=2...