a(n)是等差数列,设f（x)=a(1)x+a(2)x^2+...+a(n)x^n,n是正偶数,且已知fn(1)=n^2,fn(-1)=n(1)求数列a(n)的通项公式(2)证明5/4
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a(n)是等差数列,设f（x)=a(1)x+a(2)x^2+...+a(n)x^n,n是正偶数,且已知fn(1)=n^2,fn(-1)=n(1)求数列a(n)的通项公式(2)证明5/4
a(n)是等差数列,设f（x)=a(1)x+a(2)x^2+...+a(n)x^n,n是正偶数,且已知fn(1)=n^2,fn(-1)=n
(1)求数列a(n)的通项公式
(2)证明5/4
a(n)是等差数列,设f（x)=a(1)x+a(2)x^2+...+a(n)x^n,n是正偶数,且已知fn(1)=n^2,fn(-1)=n(1)求数列a(n)的通项公式(2)证明5/4
1.
f(x)=a1*x+a2*x^2+...+an*x^n
f(1)=a1+a2+a3+...+a(n-1)+an=n^2
f(-1)=-a1+a2-a3+...-a(n-1)+an=n
n=2时,
a1+a2=4,-a1+a2=2
a1=1,a2=3
d=3-1=2
an=1+2(n-1) =2n-1
2.
fn(x)=x+3x^2+...+(2n-3)x^(n-1)+(2n-1)x^n
xfn(x)=x^2+3x^3+...+(2n-3)x^n+(2n-1)x^(n+1)
(1-x)fn(x)=x+2x^2+...+2x^(n-1)+2x^n-(2n-1)x^(n+1)
=2x(x^n-1)/(x-1)-(2n-1)x^(n+1)-x
xfn(x)=x^2+3x^3+...+(2n-3)x^n+(2n-1)x^(n+1)
fn(x)=[2x(x^n-1)/(x-1)-(2n-1)x^(n+1)-x]/(1-x)
fn(1/2)={[(1/2)^n-1]/(-1/2)-(2n-1)(1/2)^(n+1)-1/2}/(1/2)
=-(1/2)^(n-2)-(2n-1)(1/2)^n+3
=-4(1/2)^n-(2n-1)(1/2)^n+3
=-(2n+3)(1/2)^n+3
fn(1/2)=-(2n+3)(1/2)^n+3
因-(2n+3)(1/2)^n＜0
所以fn(1/2)=-(2n+3)(1/2)^n+3 ＜3
fn(1/2)-f(n-1)(1/2)=[-(2n+3)(1/2)^n+3]-[-(2n+1)(1/2)^(n-1)+3]
=-(2n+3)(1/2)^n+(2n+1)(1/2)^(n-1)
=(2n-1)(1/2)^n
＞0
fn(1/2)单调递增
fn(1/2)＞……＞f3(1/2)＞f2(1/2)＞f1(1/2)
fn(1/2)＞……＞15/8＞5/4＞1/2
当n≥3时,15/8＜fn(1/2) ＜3
当n≥2时,5/4＜fn(1/2) ＜3
哇·