Do not use mathematical induction to prove or derive the formula of 1 square, 2 square, n square

1^2+2^2+3^2+……n^2=n(n+1)(2n+1)/6

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

So: 2 3-1 3=2*2 2+1 2-2

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

Sum of the upper (n-1) equation:

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

n 3-(n-1) 3=n 2+n(n-1)+(n-1) 2 add up.

n 3-1 = 3 (1 square + 2 square + ..)n squared)-n 2-1-(1+2+..n)

Once you've sorted out this equation, you're good to go.

There should be seven or eight ways to do it.

It can also be done with the knowledge from the university. (See the book Numerical Analysis).

This is true if n=k

i.e. 1 2 + 2 2 + 3 + ......k^2=k(k+1)(2k+1)/6

n=k+1.

1^2+2^2+3^3+……k^2+(k+1)^2

k(k+1)(2k+1)/6+(k+1)^2

k+1)[k(2k+1)+6k+6]/6

k+1)[2k^2+7k+6]/6

k+1)(k+2)(2k+3)/6

k+1)[(k+1)+1][2(k+1)+1]/6

It is also true that 1 is squared + 2 is squared + 3 is squared + ....+n squared = n(n+1)(2n+1) 6

IntroductionMathematical induction (MI) is a mathematical proof method that is usually used to prove that a given proposition holds within a range of natural numbers as a whole (or partially).

In addition to natural numbers, mathematical induction in a broad sense can also be used to prove general good-based structures, such as trees in set theory. This generalized mathematical induction method is applied to the fields of mathematical logic and computer science and is called structural induction.

In number theory, mathematical induction is a mathematical theorem that proves in a different way that any given situation is true (the first, the second, the third, and so on).

n=1 Left=1 Right=1*2*3 6=1 Left equals right True.

This is true if n=k

i.e. 1 2 + 2 2 + 3 + ......k 2 = k (k + 1) (2k + 1) 6n = k + 1.

1^2+2^2+3^3+……k^2+(k+1)^2=k(k+1)(2k+1)/6+(k+1)^2=(k+1)[k(2k+1)+6k+6]/6=(k+1)[2k^2+7k+6]/6

k+1)(k+2)(2k+3)/6

k+1)[(k+1)+1][2(k+1)+1] 6 also holds, so 1 squared + 2 squared + 3 squared +....+n squared = n(n+1)(2n+1) 6

Mathematical induction.

When n=1 and the right side of the equation = 1*2*3 6=1, the assumption is true when n=k.

1^2+2^2……+k 2=k(k+1)(2k+1) 6 is true, then n=k+1.

Left side of the equation = 1 2 + 2 2 + ......k^2+(k+1)^2=[k(k+1)(2k+1)/6]+(k+1)^2=(k+1)[2k^2+k+6(k+1)]/6=(k+1)(2k^2+7k+6)/6

k+1)(k+2)(2k+3)/6

And when n=k+1, the right side of the equation = (k+1)(k+2)(2k+3) 6, both left = right.

Therefore, this equation also holds when n=k+1.

So the equation holds when n is any positive integer.

First prove a theorem:

1x2+2x3+3x4+、、nx(n+1）=(1/3)(1*2*3-0*1*2)+(1/3)(2*3*4-1*2*3)+(1/3)(3*4*5-2*3*4)+.1/3)[n*(n+1)(n+3)-(n-1)*n*(n+1)]

1/3)[n(n+1)(n+2)-0]=n(n+1)(n+2)/3 。。There is another summation formula.

1+2+3+..n=n(n+1)/2。。。

Okay, now +

Straight to the point. 1 squared + 2 squared + 3 squared + ...+n squared = n(n+1)(2n+1) 6

Mathematical induction is a three-step process, 1, and the verification is true for n=1.

2. Assume that n=k is true.

3. Verify that n=k+1 is true.

then is true for all n.

So the steps are as follows:

1, when n=1, (1+1) 2=2 2=4, 1 2+2*1+1=4, (1+1) 2=1 2+2*1+1, is established.

2. Assuming n=k, then hunger has (k+1) 2=k 2+2k+13, when the imitation difference n=k+1, [(k+1)+1] 2=(k+2)*(k+2)=k 2+4k+4

k^2+2k+1)+(2k+2)+1

k+1)^2+2(k+1)+1

So for any positive integer n, there is (n+1) 2=n 2+2n+1 true.

It should be n>=5 when n2=5

i.e., trace sensitive K 20

So k 2 > Kyung Lee 2k + 1

So 2 k>k 2>2k+1

The posture difference is 2k+1-2 k

Mathematical induction.

When n=1 and the right side of the equation = 1*2*3 6=1, the assumption is true when n=k.

1^2+2^2……+k 2=k(k+1)(2k+1) 6 is true, then n=k+1.

Left side of the equation = 1 2 + 2 2 + ......k^2+(k+1)^2=[k(k+1)(2k+1)/6]+(k+1)^2=(k+1)[2k^2+k+6(k+1)]/6=(k+1)(2k^2+7k+6)/6

k+1)(k+2)(2k+3)/6

And when n=k+1, the right side of the equation = (k+1)(k+2)(2k+3) 6, both left = right.

Therefore, this equation also holds when n=k+1.

So the equation holds when n is any positive integer.

n=1 Left=1 square=1 Right=1(1+1)(2*1+1) 6 =1 Left=Right.

n=2 Left = 1 square + 2 square = 5 Right = 2 (2 + 1) (2*2 + 1) 6 =5 Left = Right

n=3 Left = 1 square + 2 square + 3 square = 14 Right = 3 (3 + 1) (2*3 + 1) 6 = 14 Left = right.

n=n Left = 1 square + 2 square + 3 square + ....+n square right = n(n+1)(2n+1) 6

Add left and right and subtract to get left = right.