Question

数列（an)的前n项和为Sn=n^2+n,(bn)为等比数列,且a1=b1,b2(a2-a1)=6b1 dn=√n(n+

Answer 1

an=Sn-S(n-1)=n² + n - (n-1)² - (n-1) = 2n

∴b1=a1=2

b2=b1*(a2-a1)=4

由{bn}为等比数列,容易推出

bn=2ⁿ

则 dn=2ⁿ×√n(n+1)

Dn=d1+d2+...+dn

现在来证明Dn < n×3ⁿ

用数学归纳法,当n=1时,

D1=d1=2×√2 < 1×3¹

即当n=1时,不等式成立.

假设当n=k时不等式成立,即：

Dk < k ×3^k

那么当n=k+1时,

D(k+1)=Dk+d(k+1)

< k ×3^k + 2^(k+1) × √(k+1)(k+2)

< k ×3^k + 2^(k+1) × (k+3/2) 注：√(k+1)(k+2) < k+3/2（两面平方展开很容易发现）

= k ×3^k + 2^(k+1) × k + 2^(k+1) × 3/2

= k ×3^k + 2^k × 2 × k + 2^k × 2 × 3/2

= k ×3^k + 2^k × 2k + 2^k × 3

< k ×3^k + 3^k × 2k + 3^k × 3

= 3^k × (k+2k) +3^(k+1)

= 3^k × 3k +3^(k+1)

= 3^(k+1) × k +3^(k+1)

= (k+1) × 3^(k+1)

也就是说,当n=k+1时,原式也成立.

由数学归纳法,Dn