2的n次方乘以(2n+1),这个的前n项和怎么求?
1个回答
a1=2 a2-a1=3*2^(2-1)=6 令cn=a(n+1)-an=3*2^(2n-1),则c1=a2-a1=6,cn/c(n-1)=4 cn是首项是6公比是4的等比数列 设cn的前n-1项和为s(n-1) 则s(n-1)=an-a(n-1)+.+a2-a1=an-a1=6*[1-4^(n-1)]/(1-4)=2*[4^(n-1)-1] an=2*[4^(n-1)-1]+2=2*[4^(n-1)]=2^(2n-1) 即通项是 an=2^(2n-1) Tn=b1+.+bn=2+2*2^3+3*2^5+.+n*2^(2n-1) 2^2*Tn=4Tn= 2^3+2*2^5+...+(n-1)*2^(2n-1)+n*2^(2n+1) -3Tn=Tn-4Tn=2+2^3+2^5+.+2^(2n-1)-n*2^(2n+1)=2*(1-4^n)/(1-4)-n*2^(2n+1) =[(1-3n)*2^(2n+1)-2]/3 Tn=[(3n-1)*2^(2n+1)-2]/9
