1（1）a1 = S1 = 1
S2 -2S1 = c+1
2S3 - 3S2 = 4 +2c
S1 + S3/3 = 2* S2/2
c = -5
(2) 下班回去算第一問(wèn)貌似不對(duì)頭啊~~不好意思，算錯(cuò)了，應(yīng)該是c = 1接著來(lái)：（2）因?yàn)镾(n+1) - Sn = a(n+1)所以：n a(n+1) - Sn = n^2 +n (n-1)an - S(n-1) = (n-1)^2 + (n-1)兩式相減：n a(n+1) - (n-1)an - an = 2na(n+1) - an = 2所以an是公差為2的等差數(shù)列：an = 2n-12.（1）因?yàn)椋篴(n+1)-an=3*2^(2n-1)所以：an-a(n-1)=3*2^(2n-3)...a3-a2=3*2^3a2-a1=3*2^1上述各項(xiàng)相加：an-a1=3[2^1+2^3+2^5+2^7+...+2^(2n-3)]=3*2*[2^(2n-2)-1]/(2^2-1)=2^(2n-1)-2因此：an=2^(2n-1)（2）bn=n*2^(2n-1)Bn = 1* 2^1+ 2*2^3 + 3* 2^5 +........+ n*2^(2n-1)4Bn = 1* 2^3 + 2*2^5 +.........+(n-1)2^(2n-1) + n*2^(2n+1)上述兩式相減：-3Bn = 1* 2^1 +(2^3+2^5.......+(2n-1)) - n*2^(2n+1)Bn = n*2^(2n+1)/3 - 2^(2n+1)/9 - 2/93.(1)S（n+1）=2Sn+1a（n+1）= Sn+1an= S(n-1)+1a(n+1) - an = ana(n+1)= 2an所以是等比an= 2^(n-1)(2)bn=n*an = n 2^(n-1)Tn = 1*2^0 + 2*2^1+.......+ (n-1)2^(n-2)+n 2^(n-1)2Tn = 1*2^1+........+(n-2)2^(n-2)+(n-1)2^(n-1)+n 2^n-Tn = 1 + 2+...........+2^(n-2)+2^(n-1)-n 2^nTn = (n-1)2^n +1累死我了，敲的手都酸了！