Question

高一数学数列题,设数列{an}的前n项和为Sn,S2=5,a(n+1)=an+2n+1

(1)求an的通项公式
(2)设Tn=1/a1+1/a2+...+1/an,求证nTn<=2n-1
(3)若数列{bn}满足nan,请写出{bn}的前n项和Un的公式(无需过程)
从第(2)问不会,求大神详细解答
满足bn=nan

Answer 1

a(n+1) =an+2n+1

n=1
a2 = a1+3
(a1+a2)=2a1+3
S2 =2a1+3
5=2a1+3
a1 = 1

a(n+1) =an+2n+1
a(n+1) -an = 2n+1
an-a(n-1) = 2n-1
an -a1 = 1+3+...+(2n-1)
= n^2
an = n^2 +1
ie
an =1 ; n=1
=n^2+1 ; n>=2
(2)
for n>=2

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

Tn = 1/a1+1/a2+..+1/an
= 1+ 1/(2^2+1)+...+ 1/(n^2+1)
< 1+ (1/1-1/2)+(1/2-1/3)+...+[1/(n-1) -1/n]
= 2- 1/n
nTn <= 2n -1

(3)
bn=nan
b1+b2+...+bn= n + [∑(i:1->n) i^3-i ]

n^3 -n
= n(n+1)(n+2) - 3n(n+1)
= (1/4)[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)] - [n(n+1)(n+2) - (n-1)n(n+1)]
[∑(i:1->n) i^3-i ] = (1/4)n(n+1)(n+2)(n+3) - n(n+1)(n+2)
= (1/4)n(n+1)(n+2)(n-1)

bn=nan
b1+b2+...+bn= n + [∑(i:1->n) i^3-i ]
= n + (1/4)n(n+1)(n+2)(n-1)