已知数列an的前n项和sn=n^2+pn,数列bn的前n项和Tn=2bn-1,求数列｛an*bn｝的前n项和Mn
2个回答
a1=S1=1+p
n≥2时,
an=Sn-Sn-1
=n²+pn-(n-1)²-p(n-1)
=2n-1+p
b1=T1=2b1-1
b1=1
n≥2时,
bn=Tn-Tn-1
=2bn-1-2bn-1 + 1
=2bn-2bn-1
推出bn=2bn-1
所以bn=b1×2^(n-1)=2^(n-1)
设cn=an×bn=(2n-1+p)×2^(n-1),Un为{cn}前n项和.
则2Un-Un=Un
= (1+p)×2+(3+p)×2^2+(5+p)×2^3+(7+p)×2^4+.+(2n-5+p)×2^(n-2)+(2n-3+p)×2^(n-1)+(2n-1+p)×2^n
-[(1+p)+(3+p)×2+(5+p)×2^2+(7+p)×2^3+(9+p)×2^4+.+(2n-3+p)×2^(n-2)+(2n-1+p)×2^(n-1)]
=-(1+p)-2×2-2×2^2-2×2^3-2×2^4-.-2×2^(n-2)-2×2^(n-1)+(2n-1+p)×2^n
=-(1+p)-[2^2+2^3+2^4+.+2^(n-1)]+(2n-1+p)×2^n
=-1-p-[4-2^(n+1)]/(1-2)+(2n-1+p)×2^n
=4-2^(n+1)-1-p+(2n-1+p)×2^n
=-2×2^n+(2n-1+p)×2^n+3-p
=(2n-3+p)×2^n+3-p
相关问题
