Solution 1 ^ 2 + 2 ^ 2 + 3 ^ 2 + 4 ^ 2 +... + N ^ 2 method (solution: 1 square + 2 square + 3 square... Add the sum of N square meters)

 

 

Using formula (n-1) 3
=
N3
-3n2
+ 3n-1

 

Set S3 =
13
+ 23
+ 33
+ 43
+... + N3

And S2 =
12
+ 22
+ 32
+ 42
+... + N2

And S1 = 1 + 2 + 3 + 4 +... + n

 

D:

S3-3S2 + 3s1-n = (1-1) 3
+
(2-1) 3 +
(3-1) 3
+ (4-1) 3
+... +
(N-1) 3
= S3
-N3
 

Therefore, 3S2 = 3s1 + N3
-N

Bring S1 = n (n + 1)/2 to the upper formula, which can be:

 

S2 = n (n + 1) (2n + 1)/6

 

That is, 12
+ 22
+ 32
+ 42
+... + N2
 
= N (n + 1) (2n + 1)/6

 

We can imagine that, in the same way, we can use S4 to get S3, that is, 13.
+ 23
+ 33
+ 43
+... + N3
And so on.