Square of 2 + square of 4 + square of 6 What is the square of 50 6³+7³+8³+9³+10³+11³…… +15³

Square of 2 + square of 4 + square of 6 What is the square of 50 6³+7³+8³+9³+10³+11³…… +15³

Square of 2 + square of 4 +. + 50 square = 2 ^ 2 (1 ^ 2 + 2 ^ 2 +... + 25 ^ 2) = 4 * 25 * (25 + 1) (2 * 25 + 1) / 6 = 22100 n * (n + 1) (2 * n + 1) / 6. You actually propose a square of 2 for each term, then n = 25, and then multiply the result by the cube of the first n natural numbers of 4 and the square of formula [n (n + 1) / 2] 6 & sup3; + 7 &

How much is the square of 2 plus the square of 4 plus the square of 6 plus the square of 50

Square of 2 + square of 4 +. + 50 square = 2 ^ 2 (1 ^ 2 + 2 ^ 2 +... + 25 ^ 2) = 4 * 25 * (25 + 1) (2 * 25 + 1) / 6 = 22100 n * (n + 1) (2 * n + 1) / 6. You actually propose a square of 2 for each term, then n = 25, and then multiply the result by the cube of the first n natural numbers of 4 and the square of formula [n (n + 1) / 2] 6 & # 179; + 7 &

1+12+(13+23)+(14+24+34)+… +(150+250+350+… +4850+4950)．

How to quickly calculate 1 + 2 + 3 + 4 +... + 48 + 49 + 50 + 49 + 48 +... + 3 + 2 + 1?

Add a zero to the front
0+1+2+3+4+...+48+49+50=25*50+25
49+48+...+3+2+1=24*50+25
Results 1300

Natural numbers are arranged in the following matrix. Question: what are the numbers in the first row and the eighth column? (please write the steps)
1 2 5...
4 3 6...
9 8 7...

50
The number of the first column is the square of the number of columns
The eighth column in the first row is the first number in the seventh column + 1
It's 7 * 7 + 1 = 50

As shown in the figure, ⊙ o center angle ∠ AOB = 90 °, the distance from point O to chord AB is 4, then the diameter length of ⊙ o is______ ．

As shown in the figure, the crossing point O is OC ⊥ AB, the perpendicular foot is C, ∵ ∠ AOB = 90 °, a = ∠ AOC = 45 °, OC = AC, ∵ co = 4, ∵ AC = 4, ∵ OA = ac2 + CO2 = 42, and the diameter length of ∵ o is 82

Simple algorithm of 12.5-12.5x0.2

12.5-12.5×0.2＝
12.5×1-12.5×0.2
＝12.5×（1-0.2）
＝12.5×0.8
＝10

If the lengths of the two right sides of a right triangle are 6 and 8 respectively, then the radius of its circumcircle is______ The radius of the inscribed circle is______ ．

As shown in the figure, ∵ AC = 8, BC = 6, ∵ AB = 10, ∵ circumcircle radius is 5, let the radius of inscribed circle be r, ∵ CE = CF = R, ∵ ad = AF = 8-r, BD = be = 6-r, ∵ 6-r + 8-r = 10, the solution is r = 2

Solving equation 3 (x-12-10) = X-10

3x-36-30=x-10
2x=56
x=28

Let the curve equation be x = 2T + 3 + arctant y = 2-3T + ln (1 + T ^ 2) to find the tangent equation of the curve at x = 3

First, we get the solution of T:
Let x = 3, 3 = 2T + 3 + arctant
That is, 0 = 2T + arctant
Considering that t = 0 satisfies the above formula and the function on the right side of T is strictly monotone increasing function, t = 0 is the unique solution
So the tangent point is (3,2)
The tangent slope is
dy/dx |(t=0)=(dy/dt)/(dx/dt) |(t=0)
=[-3+2t/(1+t^2)]/[2+1/(1+t^2)] |(t=0)
=-3/(2+1)=-1
So the tangent equation is
y-2=(-1)(x-3)
That is to say
x+y-5=0