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4n+2 |
3 |
(1)由题设知,N(2n)=N(n),N(2n-1)=2n-1.
S(4)=[N(1)+N(3)+N(5)+…+N(15)]+[N(2)+N(4)+N(6)+…+N(16)]
=[1+3+5+…+15]+[N(2)+N(4)+N(6)+…+N(16)]
=43+S(3)
=43+42+S(2)
=43+42+41+S(1)=86.
(2)S(n)=[1+3+5+…+(2n-1)]+[N(2)+N(4)+N(6)+…+N(2n)],
∴S(n)=4n-1+S(n-1)(n≥1),
又S1=N(1)=1,
∴S(n)=4n-1+4n-2+…+41+40+1=
4n+2
3.
点评:
本题考点: 等比数列的前n项和.
考点点评: 本题考查等比数列的性质和应用,解题时要注意等比当选列的前n项和公式、通项公式的灵活运用,注意总结规律,认真解答.
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