Sequence: 1 × 1 + 2 × 2 + 3 × 3 +. + n × n

We can get the result of n (n + 1) (2n + 1) / 6 by split term method

Let Sn be the sum of the first n terms of the sequence {α n}, Sn = kn ^ 2 + N, n ∈ n , where k is a constant
(1) Find A1 and α n (2), if for any m ∈ n , am, A2M, A4M into an equal ratio sequence (the above data are sequences), find the value of K

(1) ∵ Sn = kn ^ 2 + n; ∵ A1 = S1 = K + 1; s (n-1) = K (n-1) ^ 2 + n-1; ∵ an = SN-S (n-1) = 2kn-k + 1; (2) ∵ am = 2km-k + 1; A2M = 4km-k + 1; A4M = 8km-k + 1; for any m belongs to n *, am, A2M, A4M is a proportional sequence; ∵ (4km-k + 1) &# 178; = (2km-k + 1) * (8km-k + 1); ∵ for any m belongs to n *, am, A2M, A4M is a proportional sequence

Let B (n) = 3 / (2n + 1) * (2n-1)
Finding the sum of the first n terms of the sequence {B (n)}

It's simple. It's the sum of the split terms
bn=3/2[1/(2n-1)-1/(2n+1)]
b1=3/2(1/1-1/3)①
b2=3/2(1/3-1/5)②
b3=3/2(1/5-1/7)③
……………………
BN = 3 / 2 (1 / (2n-1) - 1 / (2n + 1)] (n) I don't know the rule. Do you see it
Sn=①+②+③+… +（n）
=3/2[1-1/(2n+1)]
=3n/(2n+1)

Factorization: (A & sup2; + 9) & sup2; - 36 - 1 + 4 (a-b) & sup2;

There are a number of boys in the school chorus, one fourth more girls than boys, and () girls,

5a/4

First simplify and then evaluate. 1 / 2 - (- 3 / 2x + 1 / 3y2) + (2x-2 / 3y2),

1/2+(-3/2x+1/3y2)-(2x-2/3y2)
Original formula = 1 / 2-2x + 2 / 3Y ^ 2-3 / 2x + 1 / 3Y ^ 2
=1/2-7/2x+y^2
Where x = - 1 / 4, y = - 1 / 2
Original formula = 1 / 2 + 7 / 8 + 1 / 4
=13/8

There are four more girls than two-thirds of the boys in a class. If there are three fewer boys and four more girls, the number of men and women is equal. How many men and women are there?
It's a formula

The number of boys = (4 + 3 + 4) / (1-2 / 3) = 33
Number of female students = 33-7 = 26
If you use a series of equations
There are x boys and y girls
2/3X+4=Y
X-3=Y+4
The solution of simultaneous equations is x = 33, y = 26

If the solution of the system of equations x-2y = 103x + y = 5K with respect to XY satisfies the condition that 2x-y is greater than 5, then K can be obtained

From Formula 1, we get x = 10 + 2Y and substitute it into formula 2
The solution of 3 (10 + 2Y) + y = 5K is y = (5k-30) / 7 x = (10K + 10) / 7
2x-y greater than 5
2(10k+10)/7-(5k-30)/7 >5
The solution is k > - 1

Number a is 25% more than number B, and number B is ()
There must be a formula!

Let B be x, then a be (1 + 25%) * X
Number B is more than number A: [(1 + 25%) * x-x] / [(1 + 25%) * x] = (1 + 25% - 1) / (1 + 25%) = 20%
Be more specific and choose mine

Sequence: 1 × 1 + 2 × 2 + 3 × 3 +. + n × n