Question

请帮忙解决这道题

请帮忙解决这道题数学数列

Answer 1

解答

追问

谢了

追答

不客气

Answer 2

1)Sn=c1+……+cn=a1+……+an+b1+……+bn

=3n(n+1)/2-2n + 4(4^n-1)/3=4^(n+1)+3n²/2-n/2-4/3

2)dn=1/(3n-2)(3n+1)=1/3[1/(3n-2)-1/(3n+1)]

所以Tn=1/3[1/1-1/4+1/4-1/7+……+1/(3n-2)-1/(3n+1)]

=1/3[1-1/(3n+1)]=n/(3n+1)

3)en=(3n-2)*4^n

Rn=(3*1-2)*4^1+……(3n-2)*4^n

4Rn=    (3*1-2)*4^2+……(3n-2)*4^(n+1)
两式相减得
3Rn=-1-3(4^2+……+4^n)+(3n-2)*4^(n+1)

=-1-[4^(n+1)-16]+(3n-2)*4^(n+1)

=3(n-1)*4^(n+1)+15

Rn=(n-1)*4^(n+1)+5

4)Pn= (-1)(3*1-2)+(-1)²(3*2-2)+……+(-1)^n(3n-2)

(-1)Pn=    (-1)²(3*1-2)+(-1)^3(3*2-3)+……+(-1)^(n+1) (3n-2)

两式相减
2Pn=(-1)(3*1-2)+3[(-1)²+(-1)^3+……+(-1)^n]-(-1)^(n+1) (3n-2)

=-1+3[1-(-1)^(n-1)]/2-(-1)^(n+1) (3n-2)

=1/2+(3n-1/2)(-1)^n

Pn=1/4+(6n-1)(-1)^n/4