由题意知a+a≠0,∴a-a=4. n+1nn+1n 即数列{a}为等差数列,其中a=2,公差d=4.∴a=a+(n-1)d=2+4(n-1), n1n1 即通项公式为a=4n-2. n (3)解:令c=b-1,则 nn ?? 1aa n+1n ?? c=+?2 n ?? 2aa ?nn+1? 1?2n+12n?1? ???? =?1+?1 ???? ?? 22n?12n+1 ???? ?? 11 =?, 2n?12n+1 b+b+…+b-n=c+c+…+c 12n12n 11111 ?????? =1?+?+?+? ?????? 3352n?12n+1 ?????? 1 =1?. 2n+1 1 ?? ∴lim(b+b+?+b?n)=lim1?=1 ?? 12n n→∞n→∞ 2n+1 ??
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