举报
pppfff0020
(1). S[n+1]=(n+1)A[n+1]-2n(n+1) (a) S[n]=nA[n]-2n(n-1) (b) (a)-(b),得: S[n+1]-S[n] =(n+1)A[n+1]-2n(n+1)-nA[n] + 2n(n-1) =(n+1)A[n+1]-nA[n]-4n =(n+1)A[n+1]-n(A[n]-4) 又,S[n+1]-S[n]=A[n+1] 故:A[n+1]=(n+1)A[n+1]-n(A[n] + 4) 即:n(A[n] + 4)=(n+1)A[n+1]-A[n+1]=nA[n+1] n>0,从而:A[n] + 4=A[n+1] 即:A[n]为公差为4的等差数列,又A[1]=1, 故,其通项公式为:A[n]=4n-3。 (2) 1/(A[n-1]A[n])=1/((4n-7)(4n-3))=(1/(4n-7)-1/(4n-3))/4=(1/A[n-1]-1/A[n])/4 因此, 1/(A[1]A[2])+1/(A[2]A[3])+……+1/(A[n-1]A[n]) =(1/A[1]-1/A[2])/4 + (1/A[2]-1/A[3])/4 + …… + (1/A[n-1]-1/A[n])/4 =(1/4)*(1/A[1]-1/A[2]+1/A[2]-1/A[3] + …… + 1/A[n-1]-1/A[n]) =(1/4)*(1/A[1]-1/A[n]) (K) A[1]=1,A[n]〉0, 1/A[1]=1,1/A[n]〉0, 因此,1/A[1]-1/A[n]〈1 从而(K)式〈1/4。 证毕。