Question

已知数列an bn的通项an bn 满足 bn=an乘2的n次方 且数列an的前n项和sn=n2次方-2n

Answer 1

数列an的前n项和Sn=n^2-2n

n=1时

a1=S1=1-2=-1

n>=2时

an=Sn-S(n-1)=(n^2-2n)-((n-1)^2-2(n-1))

=2n-3

n=1时,满足an=2n-3

∴an=2n-3

(2)

bn=an*2^n

=(2n-3)*2^n

b1=-1*2^1

b2=1*2^2

b3=3*2^3

.

bn=(2n-3)*2^n

错位相减法

Tn=b1+b2+b3+.+bn

=-1*2^1+1*2^2+3*2^3+.+(2n-3)*2^n

2Tn= -1*2^2+1*2^3+3*2^4+.+(2n-1)*2^n+(2n-3)*2^(n+1)

Tn-2Tn=-1*2^1+2*2^2+2*2^3+.+2*2^n-(2n-3)*2^(n+1)

-2Tn=-2+2(2^2+2^3+.+2^n)-(2n-3)*2^(n+1)

-2Tn=-2+2*(2^2-2^n*2)/(1-2)-(2n-3)*2^(n+1).等比数列求和公式

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

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

=5+(2n-5)*2^n

验证

n=1,b1=T1=5+(-3)*2=-1

成立