a1 n 2 13n 4 sequence... Rush 50

a1=-n 2+13n+4, the first n term sn=100, then n=sn=a1+a2+a3+.an

1^2-2^2-3^2-..n^2+13+26+39+..13n+4+4+4+..4

1^2+2^2+3^2+..n^2)+13(1+2+3+..n)+4n

1^2+2^2+3^2+..n^2)+13(1+n)n/2+4n-(1^2+2^2+3^2+..n^2)+(13n^2+13n)/2+4n

1^2+2^2+3^2+..n^2)+(13n^2+21n)/2-n(n+1)(2n+1)/6+(13n^2+21n)/2100=-n(n+1)(2n+1)/6+(13n^2+21n)/2600=-n(n+1)(2n+1)+3(13n^2+21n)600=-n(2n^2+3n+1)+3(13n^2+21n)600=-2n^3-3n^2-n+39n^2+63n600=-2n^3+36n^2+62n

300=-n^3+18n^2+31n

n^3-18n^2-31n+300=0

Let's do the math for yourself.

There is a problem with the topic.

Either that's an=-n 2+13n+4

Either a1=-1 2+13*1+4=16

Divide it into three sections, n n and 4, and find sn1 + sn2 + sn3 = 100 respectively.

a1=1,a1+2a2+3a3+..nan=(n+1)(a(n+1)) 2, so that n=1 gets: a1=2a2 2, a2=1

When n 2, a1+2a2+3a3+.n-1)a(n-1)=na(n) 2, subtract the two equations to obtain: nan=(n+1)(a(n+1)) 2 -na(n) 2,3na(n) 2=(n+1)(a(n+1)) 2,a(n+1) a(n)= 3n (n+1)( n 2), so a3 a2=3 2 3,a4 a3=3 3 4,a5 a4=3 4 5, ......

a(n) /a(n-1)= 3(n-1)/n

Multiply the above equations to obtain: a(n) a2=3 (n-2) 2 n( n 2), a(n)=3 (n-2) 2 n ( n 2), in summary: n=1, a(n)=1

n2, a(n)=3 (n-2)2 n

Solution: (1) Because a1 + 2 a2 + 3a3 + ...+nan=n+1 2 an+1(n∈n*)

So a1+2a2+3a3+....+n-1)an-1=n 2 an(n 2)--1 points).

Subtract the two formulas to give nan=n+1 2 an+1-n 2 an

So (n+1)an+1 nan = 3(n 2)--2 points).

Therefore, from the second term, the series is an proportional series with 2 as the first term and 3 as the common ratio.

So nan=2 3n-2(n2)--3 points).

Therefore an= 1, n=1 2 n 3n-2, n 2 --4 points).

2) From (1), it can be seen that when n 2n2an=2n 3n-2

When n 2, tn = 1 + 4 30 + 6 31 + ....+2n 3n - 2 ,--5 points).

3tn=3+4•31+…+2(n-1) 3n-2+2n 3n-1,--6 points).

Subtract the two formulas to give tn=1 2 +(n-1 2) 3n-1(n2)--7 points).

and t1=a1=1 also satisfies the above equation ,--8 points).

So tn=1 2 =(n-1 2)3n-1(n n*)-9 points).

Find the general formula of nan first, and then find an

Solution: When n=1, a1=1*2*3=6;

According to the title: a1 + 2 a2 + 3a3 + ...nan=n(n+1)(n+2)，a1+2a2+3a3+..nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)

Subtract the two rubbers to obtain (n+1)a(n+1)=(n+1)(n+2)(n+3)-n(n+1)(n+2), so a(n+1)=(n+2)(n+3)-n(n+2)=3(n+2), thus: an=3(n+1).

n 2) It has been tested that An (Yuanbeizhou A 2) is also suitable for the case of A1 hail shelter.

Therefore, the general term an=3(n+1)

a1=1,a1+2a2+3a3+..nan=(n+1)(a(n+1)) 2, so that n=1 gets: a1=2a2 2, a2=1

When n 2, a1+2a2+3a3+.n-1)a(n-1)=na(n) 2, subtract the two equations to obtain: nan=(n+1)(a(n+1)) 2

na(n)/2,3na(n)/2=(n+1)(a(n+1))/2，a(n+1)

a(n)=3n/(n+1)(

n 2), so a3 a2=3 2 3, a4 a3=3 3 4, a5 a4=3 4 5,......

a(n)a(n-1)=

3(n-1)/n

Multiply the above to obtain: a(n).

a2=3 (n-2) 2 is known as n(

n 2), a(n)=3 (n-2) 2 letter cover n

n 2), in summary: n = 1 slippery noisy, a(n) = 1

n2, a(n)=3 (n-2)2 n

(3n^2-2n-1)an=a1+a2+a3+..a(n-1)

3(n-1)^2-2(n-1)-1]an-1=a1+a2+a3+..a(n-2)

Subtract the two formulas. 3n^2-2n-1)an-[3(n-1)^2-2(n-1)-1]a(n-1)=a(n-1)

3n^2-2n-1)an=[3n^2-6n+3-2n+2-1+1]a(n-1)

3n^2-2n-1)an=[3n^2-8n+5]a(n-1)

3n+1)(n-1)an=[3n-5][n-1]a(n-1)

3n+1)an=(3n-5)a(n-1)

an/a(n-1)=(3n-5)/(3n+1)

an/a(n-1)=[3(n-2)+1]/(3n+1)

So, there is. a(n-1)/a(n-2)=[3(n-3)+1]/[3(n-1)+1]

a4/a3=[3x2+1]/[3x4-1]

a3/a2=[3x1+1]/[3x3-1]

Multiply. an a2 = [3x2+1][3x1+1] (3n+1)[3(n-1)+1] (Note: The denominator is only the first two denominators, and the last two are left in the numerator).

By condition (12-4-1) a2 = 1 4

a2=1/28

Substituting the above formula. an = 1/(3n+1)[3(n-1)+1]

When n = 1 then an = 1 4, within the general term formula so.

an = 1/(3n+1)[3(n-1)+1]

The first question is okan = 1 (3n+1)[3(n-1)+1].

1/3[1/[3(n-1)+1] -1/(3n+1)]

So sn = a1+a2+.an

1/3[1-1/4+1/4-1/7+..1/[3(n-1)+1] -1/(3n+1)]

1/3[1-1/(3n+1)]

n/(3n+1)

The second question ok is to verify it again, and the overall work is relatively smooth, and there should be no major problems.

It is known to be.

3n-2)n * an = sn

3n^2-8n+5)*a(n-1) = s(n-1)

Subtract to get (3n 2-2n-1) *an = (3n 2-8n+5)*a(n-1).

So (3n+1) *an = (3n-5) *a(n-1).

So an = a1 * 1 7 * 4 10 * 7 13 * 3n-8) (3n-2) *3n-5) (3n+1) = 1 (3n-2)(3n+1).

2) an = 1/3 * 1/(3n-2) -1/(3n+1) )

So sn = 1 3 * 1 - 1 4 + 1 4 - 1 7 + 1 7 - 1 10 +1/(3n-2) -1/(3n+1) ]

1/3 * 3n / (3n+1)

n/(3n+1)

Calculate the first few terms of the series, a1 = 2, a2 = 20, a3 = 56, a4 = 272, 4 n + (-2) n> = 4 n - 2 n = 2 n (2 n - 1), so 1 an < = 1 2 n (2 n - 1) = 1 (2 n - 1) - 1 2 n. So 1 a1 + 1 a2 + ...1 an<1 2+1 cavity code dry 20 + 1 56 + 1 272 + (1 (2 5-1)-1 2 5 +.

1/(2^n-1)-1/2^n)。The following estimate (1 (2 5-1)-1 2 5+...1 (2 n-1)-1 2 n), let (1 (2 5-1)-1 2 5+...

1 Mold withered (2 n-1)-1 2 n) = t, then (1 (2 5-1)-1 2 5+.1/(2^n-1)-1/2^n))<1/(2^5-2)-1/2^5+..1/(2^n-2)-1/2^n)-1/2^n)=1/2(1/(2^4-1)-1/2^4+..

1/(2^(n-1)-1)-1/2^(n-1))=1/2(1/(2^4-1)-1/2^4)+1/2(1/(2^5-1)-1/2^5+..1 (2 n-1)-1 2 n)) -1 2(1 (2 n-1)-1 2 n)) is t<1 2(1 (2 4-1)-1 2 4)+t 2-1 2(1 (2 n-1)-1 2 n))< 1 2(1 Wudong(2 4-1)-1 2 4)+t 2, t<(1 (2 4-1)-1 2 4)=1 (15x16)=1 240. So 1 a1 + 1 a2 + ...

1/an<1/2+1/20+1/56+1/272+1/240<7/12。

Transfer: 1 a1 + 1 a2 + 1 a3 + ......1/an<7/12 - 1 / 2^(n+3)+1)

Inductive Crack Slow Sail Bridge Method n = 1,2,3,4 Calculate by yourself

When n is large enough, the certificate is n+1, and only the car model needs to be certified.

1/ a(n+1) <1 / 2^(n+3)+1) -1 / 2^(n+4)+1)

Right = 2 (n+3) 2 (n+3)+1) (2 (n+4)+1) =1 2 (-n-3)+1) (2 (n+4)+1).

1 / 1+1)/(2^(n+4)+1) =1 / 2^(n+5)+2)

Left = 1 (4 n + (2) n) < 1 (4 n - 2 n).

Obviously when n is large enough, 4 n - 2 n > 2 (n+5)+2, so the left is right.

Proven by induction.

because an+2-an+1=2(an+1-an); a2-a1=2。Therefore, the number series is an equal proportional series with 2 as the first term and 2 as the common ratio. That is: an+1-an=2 n

1）s4=a5-a4+a4-a3+……a2-a1=a5-a1=a5-a1=2^4+2^3+……2=30；So a5=31

2) In the same way, sn=an+1-a1=2*(2 n-1). When n takes 6, s6=a7-a1=126;So a7=127. i.e. 127 is this sequence.

a number.