数列an满足a1=1/6 前n项和sn=1/2n(n+1)an 1.写出s1,s2,s3,s4,s5猜想sn 用数学归纳

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数列an满足a1=1/6 前n项和sn=1/2n(n+1)an 1.写出s1,s2,s3,s4,s5猜想sn 用数学归纳法证明

Sn=(1/2)n(n+1)an
S1=a1=1/6

S2=(1/2)2(2+1)a2
a1+a2= 3a2

a2=(1/2)a1=1/12
S2 = a1+a2= 1/6+1/12=1/4

S3 = (1/2)3(3+1)a3
S2+a3=6a3
a3 = (1/4)/5 = 1/20
S3=S2+a3= 1/4+1/20=3/10

S4=10a4
S3=9a4
a4=S3/9 = 1/30
S4=S3+a4= 3/10+1/30 = 1/3

S5=15a5
a5 = S4/14 = 1/42
S5= S4+a5= 1/3+1/42=5/14

let
Sn = 1/2- 1/(n+2)
n=1
LS = S1= 1/6
RS = 1/2-1/3=1/6
n=1 is true
Assume p(k) is true
ie Sk = 1/2 -1/(k+2)
for n=k+1
S(k+1) = Sk +ak
= Sk + 2Sk/[k(k+1)]
= Sk ( 1+ 2/[k(k+1)])
= Sk ( k^2+k+2)/[(k(k+1)]
= [1/2- 1/(k+2)] ( k^2+k+2)/[(k(k+1)]
= k/[2(k+2)] . ( k^2+k+2)/[(k(k+1)]
= (1/2) [ (k^2+k+2)/(k+1)(k+2) ]
= (1/2) [ (k+1)(k+2) - 2(k+1)] /(k+1)(k+2)
= (1/2)[ 1- 2/(k+2) ]
=1/2 - 1/(k+2)
p(k+1) is true
ie By principle of MI, it is true for all n

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