(②求K[5中多项式f(x)=1+x+x2+x3+x4在基1，(，(e)2,()3,()1下的坐标 3)证明∑=++ ·4由此导出数列D。=∑的通项公式 解(1)1-1 I=() x2=0+x+x(红-1)=0+()+()2 x3=x+3x(红-1)+x(x-1)(x-2)=回)+32+)3 x4=(）+7e2+6(3+()1 故所求过渡矩阵为 10000N 01111 T=00137 00016 \00001/ (2)(1,411,7,1). (③)易知任+1)+1-()+1=(k+1)().所以 ∑-∑Ie+1- 京层 x=0 中+- =+7+1)+1 (4)因为x=()+7)2+6(3+()4,所以 Dn-∑r-∑（回+7aP+63+9 =0 =号a+1y2+了m+12+8a+1+a+1 =30nn+12m+13m2+3n-1. 习题7-2 1.给定K3的两个基 c1=(1,1,-1),1=(L,-12), 62=(1,0,-1,2=(2,-1,2), e8=(1,l,1),g=(-2,1,1). 设4为K3的线性变换使： 6=hi=1,2,3. (1)求由基e1,e2,e3到基1,2，%的过渡矩阵 (②)求在基s1,e2,c3下的矩阵 3.(2)  K[x]5 (9:; f(x) = 1 + x + x 2 + x 3 + x 4  1,hxi,hxi 2 ,hxi 3 ,hxi 4 ; ∗ (3) : Xn x=0 hxi k = 1 k + 1 hn + 1i k+1; ∗ (4) XYZ6[ Dn = Xn k=0 k 4 \:];. : (1) 1 = 1 x = hxi x 2 = 0 + x + x(x − 1) = 0 + hxi + hxi 2 x 3 = x + 3x(x − 1) + x(x − 1)(x − 2) = hxi + 3hxi 2 + hxi 3 x 4 = hxi + 7hxi 2 + 6hxi 3 + hxi 4 S& T =   1 0 0 0 0 0 1 1 1 1 0 0 1 3 7 0 0 0 1 6 0 0 0 0 1   . (2) (1, 4, 11, 7, 1). (3) ^U hx + 1i k+1 − hxi k+1 = (k + 1)hxi k . &' Xn x=0 hxi k = 1 k + 1 Xn x=0 [hx + 1i k+1 − hxi k+1] = 1 k + 1 " nX +1 x=1 hxi k+1 − Xn x=0 hxi k+1# = 1 k + 1 (hn + 1i k+1 − h0i k+1) = 1 k + 1 hn + 1i k+1 . (4)  x 4 = hxi + 7hxi 2 + 6hxi 3 + hxi 4 , &' Dn = Xn x=0 x 4 = Xn x=0 ¡ hxi + 7hxi 2 + 6hxi 3 + hxi 4 ¢ = 1 2 hn + 1i 2 + 7 3 hn + 1i 3 + 6 4 hn + 1i 4 + 1 5 hn + 1i 5 = 1 30 n(n + 1)(2n + 1)(3n 2 + 3n − 1). ￾
7–2 1. ,_ K3 `: ε1 = (1, 1, −1), ε2 = (1, 0, −1), ε3 = (1, 1, 1), η1 = (1, −1, 2), η2 = (2, −1, 2), η3 = (−2, 1, 1).  A  K3  ab, c: Aεi = ηi i = 1, 2, 3. (1)  ε1, ε2, ε3  η1, η2, η3 ; (2)  A  ε1, ε2, ε3 ; · 3 ·