In the sequence an, A1 = 1, and the point [a (n), a (n-1)] (n belongs to N +) is on the image of function f (x) = x + 2 (1) The general formula for finding a (n) (2) Take 3.4.6.. 2 Λ n + 1 power + 2.. terms from the sequence a (n) to form a new sequence B (n). Try to find the general term formula BN of B (n), the first n terms and Sn

In the sequence an, A1 = 1, and the point [a (n), a (n-1)] (n belongs to N +) is on the image of function f (x) = x + 2 (1) The general formula for finding a (n) (2) Take 3.4.6.. 2 Λ n + 1 power + 2.. terms from the sequence a (n) to form a new sequence B (n). Try to find the general term formula BN of B (n), the first n terms and Sn

(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^n