12 and what number add to 12 minus what number is equal to the square of 12?

12 and what number add to 12 minus what number is equal to the square of 12?

It's 0 ~! 12 plus 0 equals 1212 minus 0 equals 12, and the square of 12 is 12

How to describe a set composed of all positive odd numbers

{x | x = 2n + 1, n > = 0 and N ∈ n}

Try to explain: for any natural number n, the value of n (n + 5) - (n-3) (n + 2) can be divided by 6

∵ n (n + 5) - (n-3) (n + 2) = (N2 + 5N) - (n2-n-6) = N2 + 5n-n2 + N + 6 = 6N + 6 = 6 (n + 1) and N ≥ 1 ∵ can always be divisible by 6

Known: 1 + 3 = 4 = 2, 1 + 3 + 5 = 9 = 3, 1 + 3 + 5 + 7 = 16 = 4, 1 + 3 + 5 + 7 + +(2n + 1) = what

When n = 1, the original formula is equal to the square of 2. When n = 2, the original formula is equal to the square of 3, so the original formula is equal to the square of (n + 1),

Given a series of numbers: 1, - 1 / 4,1 / 9, - 1 / 16, ·, according to this law, we can find: (1) the fifth number and the eleventh number (2) the 2nth number
It's best to have a formula! It's due tomorrow!

Notice that the denominator is the square of 2,3,4,5,6, and it's positive and negative
The fifth is one in 25 and the eleventh is one in 121
The 2nth number is a positive 1 / n square

Find the sum of the first n terms of the sequence whose general term formula is a * n = 2 ^ n + 2N-1

an=2^n+2n-1
It can be seen that 2 ^ n is an equal ratio sequence with the first term of 2 and the common ratio of 2
2n is the arithmetic sequence with the first term of 2 and the tolerance of 2
-1 is a constant
So the sum of the first n terms of an is transformed into the sum of an arithmetic sequence, an arithmetic sequence and a constant sequence
The sum of the first n terms of the equal ratio sequence S1 = 2 (1-2 ^ n) / (1-2) = 2 (2 ^ n-1)
Sum of the first n terms of arithmetic sequence S2 = n (2 + 2n) / 2 = n (1 + n)
The first n terms of constant column and S3 = - n
So the first n terms of an and Sn = 2 (2 ^ n-1) + n (1 + n) - n = 2 (2 ^ n-1) + n ^ 2

The general term formula of the sequence is 2n △ (n + 1), and the sum of the first n terms can be obtained
The general term formula of the sequence is 2n / N + 1, and the sum of the first n terms can be obtained

2n/n+1=(2n+2-2)/(n+1)=2-2/(n+1)
∴Sn=2n-2[1/2+1/3+1/4+..+1/(n+1)]
The next step is not to simplify
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The general formula of A1 = 3, an + 1 = 3an-2 (n ∈ n *)

a(n+1)=3a(n)-2
∴ a(n+1)-1=3a(n)-3=3[a(n)-1]
∴ [a(n+1)-1]/[a(n)-1]=3
That is, {a (n) - 1} is an equal ratio sequence, the first term is A1-1 = 2, and the common ratio is 3
∴ a(n)-1=2*3^(n-1)
∴ a(n)=1+2*3^(n-1)

The general term formula A1 = 3, a (n + 1) = 3an-2

a(n+1) -1=3an-3=3(an -1)
[a(n+1)-1]/(an -1) =3
That is to say, the sequence an - 1 is an equal ratio sequence with the common ratio of 2 and the first term of 2
an -1=2*3^(n-1)
an=1+2*3^(n-1)

The solution of general term formula of A1 = 2, a (n + 1) = (3an + 1) / (an + 3)

Let BN = an-1
a（n+1）=(3an+1)/(an+3)
a（n+1）-1=(3an+1)/(an+3)-1
B (n + 1) = 2bn / (BN + 4) (reverse)
Let FN = 1 / BN
f（n+1）=2fn+1/2
f（n+1)+1/2=2(fn+1/2)
FN + 1 / 2 is an equal ratio sequence
fn+1/2=3·2^（n-2）
therefore
an=1/[3·2^（n-2）-1/2]+1