求数列通项

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求数列通项

a(n+1) = [(4n+6)an +4n+10]/(2n+1)
(2n+1)a(n+1) = (4n+6)an +4n+10
(2n+1)[a(n+1)+2] = (4n+6)][an +2]
[a(n+1)+2]/[an +2] =(4n+6)/(2n+1)
[an+2]/[a(n-1) +2] =(4n+2)/(2n-1)
= 2 + 4/(2n-1)
(an+2)/(a1 +2) = 2(n-1) + 4[ 1/3+1/5+...+1/(2n-1) ]
an + 2 = (a+2) {2(n-1) + 4[ 1/3+1/5+...+1/(2n-1) ] }
an = -2+ (a+2) {2(n-1) + 4[ 1/3+1/5+...+1/(2n-1) ] }
再问：请问要怎么知道加2？ (2n+1)[a(n+1)+2] = (4n+6)][an +2]
再答： (2n+1)a(n+1) = (4n+6)an +4n+10 (2n+1)[a(n+1)+k] = (4n+6)[an +k] coef. of n 2k=4 k=2 (1) coef. of constant 5k=10 k=2 (2) (1)=(2) =>(2n+1)[a(n+1)+2] = (4n+6)[an +2]
再问： (2n+1)[a(n+1)+k] = (4n+6)[an +k不好意思，再麻烦你写一下这一步以后怎么做，一边是2n+1一边是4n+6怎么化简]
再答： (2n+1)a(n+1) = (4n+6)an +4n+10 (1) (2n+1)[a(n+1)+k] = (4n+6)[an +k] (2n+1)a(n+1) = (4n+6)an + (2k)n +5k (2) compare (1) and (2) coef. of n 2k =4 =>k=2 (3) coef. of constant 5k=10 => k=2 (4) (3)=(4) => k=2

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