Observe the equation 1x2 = 1 / 2 minus 1 / 2, 2x3 = 1 / 2 minus 1 / 3, 3x4 = 1 / 3 minus 1 / 4 (1) Fill in the blanks according to the above rules (2) According to the above rules, the calculation is: 1x2 / 1 + 2x3 / 1 + 3x4 / 1 +... + 2012x2013 / 1 (3) If the value of a is the smallest positive integer and B is 3, find one part of AB + (a + 2) (B + 2) + (a + 4) (B + 4) +... + (a + 200) (B + 200)

Observe the equation 1x2 = 1 / 2 minus 1 / 2, 2x3 = 1 / 2 minus 1 / 3, 3x4 = 1 / 3 minus 1 / 4 (1) Fill in the blanks according to the above rules (2) According to the above rules, the calculation is: 1x2 / 1 + 2x3 / 1 + 3x4 / 1 +... + 2012x2013 / 1 (3) If the value of a is the smallest positive integer and B is 3, find one part of AB + (a + 2) (B + 2) + (a + 4) (B + 4) +... + (a + 200) (B + 200)

1x21 = 1 minus 1 / 2, 2x1 / 3 = 1 / 2 minus 1 / 3, 3x4 = 1 / 3 minus 1 / 4... Use natural number n to express the above formula
The general law of the son!

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

In the isochromatic sequence {an} with non-zero tolerance, A2, A3 and A6 constitute three consecutive terms of the isochromatic sequence, and seek the value of common ratio Q!

A2, A3 and A6 constitute the continuous three terms of the equal ratio sequence
The square of A3 = a2a6
(a1+2d)²=（a1+d)(a1+5d)
It is reduced to d = - 2A1
q=a3/a2=(a1+2d)/(a1+d)
=(-3a1)/(-a1)
=3

In the arithmetic sequence {an}, A1 = 2, tolerance D ≠ 0, and A1, A3 and a11 are just the first three terms of an arithmetic sequence, then the common ratio of the arithmetic sequence is ()
A. 2B. 12C. 14D. 4

In the arithmetic sequence {an}, A1 = 2, A3 = 2 + 2D, a11 = 2 + 10d. Because A1, A3 and a11 are just the first three terms of an arithmetic sequence, there is A32 = a1a1a11, that is, (2 + 2D) 2 = 2 (2 + 10d), and the solution is d = 3, so the common ratio of the arithmetic sequence is 82 = 4, so D is selected

The first term of an arithmetic sequence {an} with non-zero tolerance is A1 = 2, and A1, A3 and a11 are three consecutive terms in the arithmetic sequence
Seeking common ratio Q

a3=2+2d,a11=2+10d
(2+2d)^2=2(2+10d)
d=
q=a3/a1

In the arithmetic sequence {an}, A1 = 2, tolerance D ≠ 0, and A1, A3 and a11 are just the first three terms of an arithmetic sequence, then the common ratio of the arithmetic sequence is ()
A. 2B. 12C. 14D. 4

In the arithmetic sequence {an}, A1 = 2, A3 = 2 + 2D, a11 = 2 + 10d. Because A1, A3 and a11 are just the first three terms of an arithmetic sequence, there is A32 = a1a1a11, that is, (2 + 2D) 2 = 2 (2 + 10d), and the solution is d = 3, so the common ratio of the arithmetic sequence is 82 = 4, so D is selected

It is known that the equal ratio sequence {BN} is equal ratio sequence, the sum of the first n terms is Sn, and the common ratio Q > 1, B1

Analysis:
（S11-a4） -（a11-S4）
=（S11+S4）-（a4+a11）
=（S11+S4）-（a5+a10）
Because it is an equal ratio sequence {BN}, the sum of the first n terms is Sn, and the common ratio Q > 1, B1

It is known that the sum of the first n terms of the arithmetic sequence {an} is Sn, and S10 = 55, S20 = 210

If the first term is A1 and the tolerance is D, then S10 = a1 + A2 + +a10 =10a1+45d=55 (45=1+2+ …… +9)
s20=20a1+190=210 (190=1+2+ …… +19)
The solution is A1 = 1, d = 1
So the general formula is an = n

Let Sn be the sum of the first n terms of the arithmetic sequence {an}, and tn be the product of the first n terms of the arithmetic sequence {BN}. (1) prove that the sequence S10, s20-s10, s30-s20 are arithmetic sequences, and give a more general conclusion (only the conclusion is required, and no proof is necessary); (2) if T10 = 10, T20 = 20, find the value of T30? Analogy (1) what conclusion can you draw? It is only required to give a conclusion, not to prove it

(1) It is proved that: let the tolerance of the arithmetic sequence {an} be D, then Sn = Na1 + n (n − 1) 2D, so S10 = 10A1 + 10 × 92d = 10A1 + 45d. Similarly, S20 = 20a1 + 190D, S30 = 30a1 + 435d. So, s20-s10 = 10A1 + 145d, s30-s20 = 10A1 + 245D, so S10 + (s30-s20) = 20a1 + 290D =

The sum of the first n terms of the arithmetic sequence {an} is denoted as SN. It is known that A2 = 1, S10 = - 25, if BN = (an) ^ 2 - [(an) + 1] ^ 2
Finding the first n terms and TN of sequence {BN}

The equations of a1 + q = 1 and 10A1 + 10 × (10-1) / 2 · q = - 25 are solved. When A1 = 2, q = - 1 reduces BN = (an) ^ 2 - (an) ^ 2-2an-1 = - 2 · [2 + (n-1) · (- 1)] - 1 = 2n-7b1 = - 5, that is, when BN ≤ 0, TN = | n / 2 · (- 5 + 2n-7)] | = | n (n-6) | = n