Cn=[(n+4)(n+5)]/[(n+1)(n+2)].Sn为数列｛Cn｝的前n项和,证明Sn

cn=(n+4)(n+5)/[(n+1)(n+2)]
= 1 + 6(n+3)/(n+1)(n+2)
= 1+6[2/(n+1) -1/(n+2) ]
Sn= c1+c2+..+cn
= 1+6[1/2 -1/(n+2)] + { summation(i:1->n) 1/(i+1) }
= 4 -1/(n+2)+ { summation(i:1->n) 1/(i+1) }
n=1
L.S=S1=c1 = 30/6=5
R.S=1+6(1+ln1) =7>L.S
p(1) is true
Assume p(k) is true
ie
4 -1/(k+2)+ { summation(i:1->k) 1/(i+1) } for n=k+1
L.S
=S(k+1)
=Sk + c(k+1)
= 4 -1/(k+2)+ { summation(i:1->k) 1/(i+1) } + (k+5)(k+6)/[(k+2)(k+3)]
=k+6(1+lnk) + 1+ (6k+24)/[(k+2)(k+3)]
=(k+1)+6(1+lnk) +6[2/(k+2) -1/(k+3)]
=(k+1) +6(1 + [ lnk+2/(k+2) - 1/(k+3)] )