The sequence a n a n n 2 a1 1 a2 4 finds a n and S n

The easiest way to do this is to put 1, 2, and 3 numbers into n.

Reasoning the words of pushing:

Below I've written a, subscript, n, etc. as an.

a(n+1)=an+2n then an=a(n-1)+2*(n-1) a(n-1)=a(n-2)+2(n-2).

In the end, a2=a1+2 doesn't add it all up, and on the left side of the equal sign is the sum of a(n+1) to a2.

The right side is the sum of an to a1 plus 2n (n from 1 n) and then the left side cancels out the right side to get a(n+1)=a1+2*(1+2+..n)

So a(n+1)=1+2*(1+2+..n) substitution n=2

Then a3=1+2*(1+2)=7

a(n)-a(n-2)=2

a(n): a(n)=n when n is an odd number

When n is an even number, a(n)=n+2

s(n): when n is odd, s(n)=n(n+1) 2+(n-1)n, when n is even, s(n)=n(n+1) 2+(n+1).

a(n)=n+1+(-1)^n

s(n)=n(n+1) 2+n+(-1) n upstairs is too rough, right? The correctness hasn't been tested, but the expressions are still segmented with parity, are you from junior high school?

1) n is a very basic knowledge point.

I'm not a math student, but I just want to ask a question, since it's a sequence.

There should be a difference between equality and proportionality.

Because a 1 =1 a 2 =3 a 3 =2 a 4 =a 3 -a 2 =-1 a 5 =-3 a 6 =-2 a 7 =1 a 8 =3, so the easy-to-know sequence is a series with 6 as the weekly shouting period, s 6 =0, and the first type of the person is 2 007 =s 364 6+3 =364s 6 +(a 1 +a 2 +a 3 )=6

a(n+2)=3a(n+1)-2ana(n+2)-a(n+1)=2a(n+1)-2an=2[a(n+1)-an][a(n+2)-a(n+1)]=2[a(n+1)-an]a2-a1=1-1=0, the number chain is a constant series of numbers with 0 terms for each shed. a(n+1)=an and a1=1, so the general term of an=1 is an=1

Solution: A(n+1)=2an+2n-3 is simplified to obtain a(n+1)+2(n+1)-1=2(an+2n-1), so it can be regarded as the first term is a1+2*1-1=3, and the common ratio is 2 in the proportional series an+2n-1=(a1+2*1-1)*2 (n-1)=3*2 (n-1).

i.e. an=3*2 (n-1)-2n+1

a(n+1)=2an+2n-3

a(n+1)+2(n+1)-1=2(an+2n-1) makes bn=an+2n-1

b1=2+2-1=3

bn=3*(2^n-1)=an+2n-1

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

Probably, I dare not guarantee whether there is anything wrong or not.

a(n+1)=2a(n) [a(n)+2]The number of inverted laughs is:

1 a(n+1)=[a(n)+2] slow roll(2a(n))=1 2+1 a(n).

1/a(n+1)-1/a(n)=1/2

The number column is an equal difference series, and the remainder is d=1 2

1/an=1/a1+(n-1)/2=(n+1)/2an=2/(n+1)

a(n+2)=2a(n+1)-a(n)+2a(n+2)-a(n+1)=a(n+1)-a(n)+2 so that b(n+1)=a(n+2)-a(n+1) then there is: b(n+1)-bn=2

The number column is a blind series of equal difference of Kenai, d=2, b1=a2-a1=3bn=3+2(n-1)=2n+1

That is: a(n+1)-a(n)=2n+1

a(n)-a(n-1)=2n-1

a(n-1)-a(n-2)=2n-3

a2-a1=3

n formulas are added left and right:

a(n+1)-a1=3+5+…Staring at Shenchun....+(2n+1)=n(n+1)a(n+1)=n(n+1)+1

an=n(n-1)+1

a（n+1）+2an=0

a（n+1）=-2an

a(n+1)/an=-2

So the sequence an is a proportional series with a common ratio of -2, so an=a1q (n-1)=2*(-2) (q-1)=2*(-2)(-2) n=-4*(-2) n

sn=a1(q^n-1)/(q-1)

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

a（n+1）+2an=0

a（n+1）=-2an

So it's a proportional series, q=-2

a1=2, so an=2*(-2) (n-1).

(-2)*(2)^(n-1)

i.e. an=-(-2) n

sn=a1*(1-q^n)/(1-q)

2*[1-(-2)^n]/(1+2)

2/3-2*(-2)^(n)/3

2/3+(-2)^(n+1)/3

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

a(n+1)-1=2a(n)-2=2(a(n)-1). a(n+1)-1) (a(n)-1)=2 then the series is proportional and the common ratio is 2

Rule. a(n)-1=(a(1)-1)q^(n-1)=(1-1)*2^(n-1)=0

a(n)=1

Because a(n+1)=2a(n)-1, a2=2a1-1, a2=1a3=2a2-1=1

a4=2a3-1=1

It is not difficult to see that the relation of this series is an=1

Solution: 2a2=a1+a3, we get 2*(a+2)=a+2a-2, a=6 So the general formula an=a1+(n-1)d=6+(n-1)*2=2n+4 is a3-a2=a2-a1 from the series of equal differences