数列〔an〕满足an+1+an=4n-3,当a1=2时,Sn为数列〔an〕前n项和,求S 2n+1

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数列〔an〕满足an+1+an=4n-3,当a1=2时,Sn为数列〔an〕前n项和,求S 2n+1
数列〔an〕满足an+1+an=4n-3,当a1=2时,Sn为数列〔an〕前n项和,求S 2n+1

数列〔an〕满足an+1+an=4n-3,当a1=2时,Sn为数列〔an〕前n项和,求S 2n+1
a(n+1) + an = 4n -3
a(n+1) - 2(n+1) + 5/2 = - ( an - 2n + 5/2 )
令 bn = an - 2n + 5/2
b1 = a1 - 2 + 5/2 = 5/2
b(n+1) = - bn
∴bn = 5/2 *(-1)^(n+1)
an = bn +2n - 5/2 = 5/2 [ (-1)^(n+1) - 1 ] + 2n
当n为奇数时,an = 2n
当n为偶数时,an = 2n - 5
当n为奇数时,Sn = (2+4+6+……+2n) - (n-1)/2 *5 = n² - 3/2 n + 5/2
当n为偶数时,Sn = (2+4+6+……+2n) - n/2 *5 = n² - 3/2 n
可合并为 Sn = n² - 3/2 n + 5/4 * [ (-1)^(n+1) + 1]
2n+1为奇数
S(2n+1) = (2n+1)² - 3/2 (2n+1) + 5/2
= 4n² - 7n + 2

分奇偶

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