1+2+3+4+5+6… ... + 99 with simple calculation

1+2+3+4+5+6… ... + 99 with simple calculation

five thousand and fifty

By using a simple method, it shows that 3 ^ 200-4 * 3 ^ 199 + 10 * 3 ^ 198 can be divisible
Explain the process and reason step by step, and how this step came

3^200-4×3^199+10×3^198
=(3²-4×3+10)×3^198
=(9-12+10)×3^198
=7×3^198
∵ 7 × 3 ^ 198 can be divided by 7
3 ^ 200-4 × 3 ^ 199 + 10 × 3 ^ 198 can be divided by 7

-Factorization method for solving inequality in x-square-2x + 3

The original formula - X & # 178; - 2x + 3
＝﹣﹙x²＋2x－3﹚
＝﹣（x＋3）（x－1）
The original formula is a factor, which can only be decomposed here. There is no inequality

5 / 6 times [(3 and 1 / 3 minus 1.2) divided by 29% times 3 / 4-0.6]

5 / 6 times [(3 and 1 / 3 minus 1.4) △ 29% times 3 / 4-0.6]
=5 / 6 times [29 / 15 △ 29% times 3 / 4-0.6]
=5 / 6 times [20 / 3 times 3 / 4-0.6]
=5 / 6 times [5-0.6]
=5 / 6 times 4.4
=11 / 3
=Three and two thirds

Saw a 20 DM long cuboid wood into three sections, the surface area increased by 28 DM2. What's the volume of this cuboid wood

Saw into 3 sections, added 4 faces. (3-1) * 2 = 4
28 / 4 * 20 = 140 cubic decimeter

6 divided by () = 1 / 16 () = 0.75 = 75% = 21: () = () fold

6 divided by (8) = 1 / 16 (12) = 0.75 = 75% = 21: (28) = (75)

Prove 1 & # 179; + 2 & # 179; + 3 & # 179; +. + n & # 179; = {1 / 2n (n + 1)} & # 178;

(1) When n = 1, the equation holds. (2) suppose n = k, the equation holds, that is, 1 ^ 3 + 2 ^ 3 + 3 ^ 3 + 4 ^ 3 + +k^3=(1+2+3+…… +k) When n = K + 1, the left side of the equation is 1 ^ 3 + 2 ^ 3 + 3 ^ 3 + 4 ^ 3 + +k^3+(k+1)^3=1+2+3+…… +k)^2+(k+1)^3=1+2+3+…… +k)^2+(k+1...

29 30 31 32 33 34 35 36 37 fill in the sum of the three numbers in the nine palace, which are equal to 99?

32 37 30
31 33 35
36 29 34

(a+b)²=a²+2ab+b²
Describe the formula in your own words

The first square, the last square, the product of the first and the last two times in the center!

Put the nine numbers - 4, - 3, - 2, - 1,0,1,2,3,4 into the nine palace lattice respectively, so that the three numbers on each row, column and diagonal can be added to get 5
Each number cannot be repeated
Just

It's impossible
If each line is 5, then the sum of the numbers in the whole Jiugong grid should be 5 * 3 = 15, but the sum of the nine numbers you give is 0, so it's impossible