Find the square of 1, the square of 2, the square of 3, and the square of n

This can be considered a formula, which should be remembered, and the derivation process is as follows:

n+1)³=n³+3n²+3n+1

n =(n-1) +3(n-1) +3(n-1)+1n-1) =(n-2) +3(n-2) +3(n-2)+1 Add these equations to get (n+1) =3 n +3 n+(n+1)n=n(n+1) 2 This is fine, simplify it to get n =n(n+1)(2n+1) 6

ps: Using (n+1) 4=n 4+4n +6n +4n+1, we can derive n =n (n+1) 4 Of course, n is used here

Practical this method can be used to launch any n k, as long as you know in advance: n (k-1), n (k-2), n (k-3) ......n, n is fine, but the amount of calculation will be larger and larger, generally remember that n or less is fine, and then the high-level ones will generally only require you to use mathematical induction to verify.

Make use of the cubic variance formula.

n^3-(n-1)^3=1*[n^2+(n-1)^2+n(n-1)]

n^2+(n-1)^2+n^2-n

2*n^2+(n-1)^2-n

n^3-(n-1)^3=2*n^2+(n-1)^2-n

The equations are added in full.

n^3-1^3=2*(2^2+3^2+..n^2)+[1^2+2^2+..n-1)^2]-(2+3+4+..n)

n^3-1=2*(1^2+2^2+3^2+..n^2)-2+[1^2+2^2+..n-1)^2+n^2]-n^2-(2+3+4+..n)

n^3-1=3*(1^2+2^2+3^2+..n^2)-2-n^2-(1+2+3+..n)+1

n^3-1=3(1^2+2^2+..n^2)-1-n^2-n(n+1)/2

3(1^2+2^2+..n^2)=n^3+n^2+n(n+1)/2=(n/2)(2n^2+2n+n+1)

n/2)(n+1)(2n+1)

1^2+2^2+3^2+..n^2=n(n+1)(2n+1)/6

1 square + 2 square + 3 square + 4 square +...n squared = n(n+1)(2n+1) 6

1 2 = 1, 2 2 = 4, 3 2 = 9, liter touch 4 2 = 16

The rest of the part is shown in the figure below

1^2+2^2+3^2+…+n^2=n(n+1)(2n+1)/6

Proof: (using the identity (n+1) 3=n 3+3n 2+3n+1):

n+1)^3-n^3=3n^2+3n+1,n^3-(n-1)^3=3(n-1)^2+3(n-1)+1

Add the two ends of this n equation separately to get:

n+1)^3-1=3(1^2+2^2+3^2+..n^2)+3(1+2+3+..n)+n, due to 1+2+3+.n=(n+1)n 2, the above formula gets:

n^3+3n^2+3n=3(1^2+2^2+3^2+..n^2)+3(n+1)n/2+n

a^2+b^2=a(a+b)-b(a-b)

Upstairs is the correct solution. Derivation process:

Mathematical induction).

1 squared + 2 squared.

Hypothesis. 1 squared + 2 squared + 3 squared + ......n squared.

n(n+1)(2n+1)/6

So. 1 squared + 2 squared + 3 squared + ......The square of n + the square of (n + 1).

n(n+1)(2n+1)/6

n+1)*(n+1)

n+1)/6

n(2n+1)+6(n+1))

n+1)((n+1)+1)(2(n+1)+1) 6 Therefore, the assumption holds. Finish.

1²+2²+…n n(n+1)(2n+1) 6 I only know how to use mathematical induction. Proof that: 1) when n 1 left 1, right 1 (1 1) (2 1) 6=1 left right The equation holds 2) When n k the equation holds i.e. 1 +2 +....k k(k+1)(2k+1) 6 n k 1 at 1 +2 +....k (k 1) k(k+1)(2k+1) 6 (k 1) k(k+1)(2k+1) 6 6(k 1) 6= 6=(k+1) [2k +7k+6] 6=(k+1)(k+2)(2k+3) 6=(k+1)[(k+1)+1][2(k+1)+1] 6, i.e., the equation holds when n k 1.

Questions. My question is **yes**.

Will you do it wow, I spent a lot of money on me, this is the topic of freshman calculus, will you do wow, please, brother.

I'm sorry for your trouble, please ask a professional mathematician, I'm sorry I can't answer you.

Questions. Then I've paid the money, I'm going to ask who, I'm going to ask you, why are you like this?

I'm really sorry, dear.

The sum of squares formula n(n+1)(2n+1) 6 is 1 2+2 2+3 2+....+n 2=n(n+1)(2n+1) 6 (Note: the square of the cover n 2=n) proves that 1 round muffled shouting 4 9 ....n 2 n(n+1)(2n+1) 6 Proofing 1 (inductive guessing ideas): 1. When Lapye n 1, 1 1(1 1)(2 1 1) 6 1 2, n 2, 1 4 2(2 1)...

Summary. Start by calculating the exponential operation: -2 2 = 1 * 2 2 = 1 * 4 = 4.

Then the division operation is calculated: 27 (-3) = 9. Finally, add the results together:

2 squared + (-1) + 27 (-3).

2 2-1+27 (-3), is that so, or (-2) 2-1+27 (-3).

A 2 2 + (-1) + 27 (-3).

First of all, the remainchain calculates the exponential operation: -2 2 = 1 * 2 2 = 1 * 4 = 4. Then calculate the division operation:

27/(-3) =9。In the end, Xuqing adds up the results and erects Sun: -4 - 1 + 9) =4 - 1 - 9 = 14.

First calculate the exponential operation: -1 2020 = 1 (the judgment of negative numbers is a positive number, such as the power of the number of numbers), and then calculate the power operation: (-2) 3 = 8, and then calculate the absolute value:

2-5|=3 Continue multiplication and dig answer: 6 * 1 2 - 13) =6 * 25 2) =75 Finally, add 1 + 8) +3 + 75) =79 so the final result is -79.

Upstairs is the correct solution. Derivation process:

Mathematical induction).

1 square + 2 square ridges.

Hypothesis. The square of 1 + 2 refers to the square of the segment + the square of 3 + ......n squared.

n(n+1)(2n+1)/6

So. 1 is squared + 2 is squared + 3 is squared + ......The square of n + the square of (n + 1).

n(n+1)(2n+1)/6

n+1)*(n+1)

n+1)/6

n(2n+1)+6(n+1))

n+1)((n+1)+1)(2(n+1)+1) 6 Therefore, the assumption holds. Finish.

=-2 for spring auction, +2=0,1- =3,3- =5

2) -1- )3- ) Substitute into the wild Jane of the Envy Hall.

+2) square foot-(1-)3-).

+2) Friendship Sleepiness - (1- )3- )

Good old +4 +4-3+4 -

Sum of Squares Formula - Title.

The sum of squares formula n(n+1)(2n+1) 6

i.e. 1 2 + 2 2 + 3 2+....+n 2 = n (n + 1) (2n + 1) 6 (Note: n 2 = n squared).

Sum of Squares Formula - Proof.

Proof 1 4 9 ....＋n^2＝n(n+1)(2n+1)/6

1, n 1, 1 1 (1 1) (2 1 1) 6 1

2, n 2, 1 4 2 (2 1) (2 2 1) 6 5

3. When n x is set, the formula holds, i.e., 1 4 9 ....＋x2＝x(x+1)(2x+1)/6

Then when n x 1, 1 4 9 ....＋x2＋（x＋1）2＝x(x+1)(2x+1)/6＋（x＋1）2

x＋1）【2（x2）＋x＋6（x＋1）】/6

x＋1）【2（x2）＋7x＋6】/6

x＋1）（2x＋3）（x＋2）/6

x＋1）【（x＋1）＋1】【2（x＋1）+1】/6

Also satisfies the formula.

4. To sum up, the formula of the sum of squares is 1 2 + 2 2 + 3 2 +....+n 2=n(n+1)(2n+1) 6 is valid, proven.