Get s [n+1] = (n+1) a [n+1]-2n (n+1) ②.

②-① a[n+ 1]-a[n]=4 indicates that the sequence {a [n]} is arithmetic progression.

If a[ 1]= 1 d=4, then a[n]=4n-3.

s[n]=n×(a[ 1]+a[n])/2=2n^2-n

(2) Let b[n]=S[n]/n=2n- 1.

Then b [1]+b [2]+…+b [n] = n× (1+2n-1)/2 = n 2 = 400.

The solution is n=20.

16 (I), from the topic1+a [2] = 2λ+1①1a [2]+4 = 2λ (1a [2])+/kloc-0.

λ= 1 is ① ② at the same time.

(ⅱ)、S[n+ 1]= 2S[n]+ 1③S[n]= 2S[n- 1]+ 1④

③-④ a[n]=2a[n- 1], that is, a[n] is a geometric series.

a[n]=2^(n- 1)

(ⅲ)、t[n]= 1×2^0+2×2^ 1+3×2^2+…+n×2^(n- 1)⑤

2t[n]= 1×2^ 1+2×2^2 +…+(n- 1)×2^(n- 1)+ n×2^n⑥

⑤-t[n]= 1×2 0+2 1+2 2+…+2(n- 1)-n×2n。

T[n]=(n- 1)2^n+ 1

t[n]/2-s[n]=(n- 1)2^(n- 1)+ 1/2-(2^n- 1)

When n≤2, t [n]/2 < s [n]

When n≥3, t [n]/2 > s [n]

15、f(x)=(x- 1)^？ Sorry, I can't see clearly.

18 ( 1)、a[ 1]=2、a[2]=2+c、a[3]=a[2]+2c=3c+2

From a[ 1], a[2], a[3] to geometric series.

The solution of 2× (3c+2) = (2+c) 2 is obtained, and the solution of c=2 is obtained.

(2)、a[n] - a[n- 1]=2(n- 1)

a[n- 1] - a[n-2]=2(n-2)

a[n-2] - a[n-3]=2(n-3)

…

a[2] - a[ 1]=2× 1

Accumulate a[n]-a[ 1]= n(n- 1)

a[n] =2 + n(n- 1)

(3)c)/(n×c^n)=(n- 1)/2^n

t[n- 1]=(n- 1)/2^n+(n-2)/2^(n- 1)+……+(2- 1)/2^2+( 1- 1)/2^ 1①

0.5t[n]=(n- 1)/2^(n+ 1)+(n-2)/2^n+……+(2- 1)/2^3+( 1- 1)/2^2②

①-②0.5t[n]= 1/2n+ 1/2(n- 1)+…+ 1/2 3+ 1/2 2-(n-65438

t[n]= 1/2+ 1/2^2+ 1/2^3+…+ 1/2^(n- 1)-(n- 1)/2^n

= 1-(n+ 1)/2^n