Question

数列的问题

帮忙..

Answer 1

S(n+1)-S(n)=a(n+1)
4a(n)-4a(n-1)=a(n+1)
2a(n)-4a(n-1)=a(n+1)-2a(n)
2b(n-1)=b(n)所以{bn}是等比数列，公比为2
a1=1,a2=S2-a1=5
b(1)=a2-2a1=3,bn=3×2^(n-1)
b(n) =a(n+1)-2a(n) =3×2^(n-1)
2b(n-1)=2[a(n)-2a(n-1)] =3×2^(n-1)
4b(n-2)=4[a(n-1)-2a(n-2)] =3×2^(n-1)
...........
2^(n-1) b(1)= 2^(n-1) [a2-2a1]=3×2^(n-1)
连续相加得到a(n+1)- 2^n ×a1 =3n× 2^(n-1)
a(n+1)=2^(n-1) ×(3n+2)
an= (3n-1)× 2^(n-2)