求解答一下第一问，，过程
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(Ⅰ)
Sn=3n²+8n
S1=3+8=11 则a1=11
S2=3*4+8*2=12+16=28
则：a2=S2-S1=28-11=17
a1=b1+b2=11
a2=b2+b3=17
设bn=b1+（n-1）d
则：b2=b1+d,b3=b1+2d
则：a1=b1+b2=b1+b1+d=11
2*b1+d=11 ①
a2=b2+b3=b1+d+b1+2d=17
2*b1+3d=17 ②
②-①:2d=6 d=3
b1=4
数列{bn}的通项公式为：bn=4+(n-1)*3=4+3n-3=3n+1
(Ⅱ)
an=bn+b(n+1)=3n+1+3(n+1)+1=6n+5
cn=(an+1)∧(n+1)/(bn+2)∧n=(6n+5+1)∧(n+1)/(3n+1+2)∧n
=(6n+6)∧(n+1)/(3n+3)∧n
=[6∧(n+1)]*[(n+1)∧(n+1)]/[3ⁿ(n+1)ⁿ]
=[2∧(n+1)][3∧(n+1)][(n+1)∧(n+1)]/[3ⁿ(n+1)ⁿ]
=[2∧(n+1)]*3*(n+1)
=3(n+1)*[2∧(n+1)]
Sn=3n²+8n
S1=3+8=11 则a1=11
S2=3*4+8*2=12+16=28
则：a2=S2-S1=28-11=17
a1=b1+b2=11
a2=b2+b3=17
设bn=b1+（n-1）d
则：b2=b1+d,b3=b1+2d
则：a1=b1+b2=b1+b1+d=11
2*b1+d=11 ①
a2=b2+b3=b1+d+b1+2d=17
2*b1+3d=17 ②
②-①:2d=6 d=3
b1=4
数列{bn}的通项公式为：bn=4+(n-1)*3=4+3n-3=3n+1
(Ⅱ)
an=bn+b(n+1)=3n+1+3(n+1)+1=6n+5
cn=(an+1)∧(n+1)/(bn+2)∧n=(6n+5+1)∧(n+1)/(3n+1+2)∧n
=(6n+6)∧(n+1)/(3n+3)∧n
=[6∧(n+1)]*[(n+1)∧(n+1)]/[3ⁿ(n+1)ⁿ]
=[2∧(n+1)][3∧(n+1)][(n+1)∧(n+1)]/[3ⁿ(n+1)ⁿ]
=[2∧(n+1)]*3*(n+1)
=3(n+1)*[2∧(n+1)]
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