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对于给定的正整数k,若数列{an}满足:an-k+an-k+1+…+an-1+an+1+…+an+k-1+an+k=2kan对任意正整数n(n>k)总成立,则称数列{an}是“P(k)数列”.(1)证明:等差数列{an}是“P(3)数列”;(2)若
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对于给定的正整数k,若数列{an}满足:an-k+an-k+1+…+an-1+an+1+…+an+k-1+an+k=2kan对任意正整数n(n>k)总成立,则称数列{an}是“P(k)数列”.
(1)证明:等差数列{an}是“P(3)数列”;
(2)若数列{an}既是“P(2)数列”,又是“P(3)数列”,证明:{an}是等差数列.
(1)证明:等差数列{an}是“P(3)数列”;
(2)若数列{an}既是“P(2)数列”,又是“P(3)数列”,证明:{an}是等差数列.
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答案和解析
(1)证明:设等差数列{an}首项为a1,公差为d,则an=a1+(n-1)d,
则an-3+an-2+an-1+an+1+an+2+an+3,
=(an-3+an+3)+(an-2+an+2)+(an-1+an+1),
=2an+2an+2an,
=2×3an,
∴等差数列{an}是“P(3)数列”;
(2)证明:当n≥4时,因为数列{an}是P(3)数列,则an-3+an-2+an-1+an+1+an+2+an+3=6an,①,
因为数列{an}是“P(2)数列”,所以an-3+an-3+an+an+1=4an-1,②,
an-1+an+an+2+an+3=4an+1,③,
②+③-①,得2an=4an-1+4an+1-6an,即2an=an-1+an+1,(n≥4),
因此n≥4从第3项起为等差数列,设公差为d,注意到a2+a3+a5+a6=4a4,
所以a2=4a4-a3-a5-a6=4(a3+d)-a3-(a3+2d)-(a3+3d)=a3-d,
因为a1+a2+a4+a5=4a3,所以a1=4a3-a2-a4-a5=4(a2+d)-a2-(a2+2d)-(a2+3d)=a2-d,
也即前3项满足等差数列的通项公式,
所以{an}为等差数列.
则an-3+an-2+an-1+an+1+an+2+an+3,
=(an-3+an+3)+(an-2+an+2)+(an-1+an+1),
=2an+2an+2an,
=2×3an,
∴等差数列{an}是“P(3)数列”;
(2)证明:当n≥4时,因为数列{an}是P(3)数列,则an-3+an-2+an-1+an+1+an+2+an+3=6an,①,
因为数列{an}是“P(2)数列”,所以an-3+an-3+an+an+1=4an-1,②,
an-1+an+an+2+an+3=4an+1,③,
②+③-①,得2an=4an-1+4an+1-6an,即2an=an-1+an+1,(n≥4),
因此n≥4从第3项起为等差数列,设公差为d,注意到a2+a3+a5+a6=4a4,
所以a2=4a4-a3-a5-a6=4(a3+d)-a3-(a3+2d)-(a3+3d)=a3-d,
因为a1+a2+a4+a5=4a3,所以a1=4a3-a2-a4-a5=4(a2+d)-a2-(a2+2d)-(a2+3d)=a2-d,
也即前3项满足等差数列的通项公式,
所以{an}为等差数列.
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