数列an中a1=1,且点【a（n）,a（n-1）】（n属于n+）在函数f（x）=x+2的图像上

(1)
f(x) =x+2
a(n-1) = an+2
an-a(n-1) =-2
an - a1 = -2(n-1)
an = -2n+3
(2)
bn = (2+2^(n+1)) an
= (2+2^(n+1)) (-2n+3)
= -4n+6 - n.2^(n+2) + 3.2^(n+1)
consider
1+x+x^2+..+x^n = (x^(n+1) -1) /(x-1)
1+2x+..+nx^(n-1)
= [(x^(n+1) -1) /(x-1)]'
= { nx^(n+1) - (n+1)x^n +1 } /(x-1)^2
multiply both side by x^2
x^2+ 2x^3 +...+ n.x^(n+1) = x^2{ nx^(n+1) - (n+1)x^n +1 } /(x-1)^2
put x = 2
1.2^2+2(2)^3+..+n(x^(n+1))
=4{ n.2^(n+1) - (n+1)2^n +1 }
bn = -4n+6 - n.2^(n+2) + 3.2^(n+1)
Sn = b1+b2+..+bn
= n(-2n+4) - 4{ n.2^(n+1) - (n+1)2^n +1 } + 12(2^n-1)
= n(-2n+4) +2^n.(12-8n+4n+4)-16
= -2n^2+4n-16 + (-4n+16) .2^nmultiply both side by x^2x^2+ 2x^3 +...+ n.x^(n+1) = x^2{ nx^(n+1) - (n+1)x^n +1 } /(x-1)^2put x = 2怎回事看不懂谢谢这是求不好意思summation（ n. 2^(n+2) ）的方法1+2x+..+nx^(n-1)= { nx^(n+1) - (n+1)x^n +1 } /(x-1)^2 x^3{1+2x+..+nx^(n-1)}= x^3 [ { nx^(n+1) - (n+1)x^n +1 } /(x-1)^2]x^3+ 2x^4 +...+ n.x^(n+2) = x^3{ nx^(n+1) - (n+1)x^n +1 } /(x-1)^2 put x=2summation（ n. 2^(n+2) ） = 8{ n.2^(n+1) - (n+1)2^n +1 } bn = -4n+6 - n. 2^(n+2) + 3. 2^(n+1)Sn = b1+b2+..+bn = n(-2n+4) - 8{ n.2^(n+1) - (n+1)2^n +1 } + 12(2^n-1)= -2n^2+4n-20 + (-8n+20). 2^n