1? +2? +……+n？ =n(n+ 1)(2n+ 1)/6 .

Proof process:

According to the cubic difference formula (a+ 1)? -a？ =3a？ +3a+ 1, with:

When a= 1: 2? - 1? =3× 1? +3× 1+ 1

When a=2: 3? -2? =3×2? +3×2+ 1

When a=3: 4? -3? =3×3? +3×3+ 1

When a=4: 5? -4? =3×4? +3×4+ 1.

When a=n: (n+ 1)? -n？ =3×n？ +3×n+ 1

Add the two sides of the equation:

(n+ 1)？ - 1=3( 1? +2? +3? + +n？ )+3( 1+2+3 ++ n)+( 1+ 1+ 1 ++ 1)

3( 1? +2? +3? + +n？ )=(n+ 1)？ - 1-3( 1+2+3+.+n)-( 1+ 1+ 1+。 + 1)

3( 1? +2? +3? + +n？ )=(n+ 1)？ - 1-3( 1+n)×n÷2-n

6( 1? +2? +3? + +n？ )=2(n+ 1)？ -3n( 1+n)-2(n+ 1)=(n+ 1)[2(n+ 1)？ -3n-2]

=(n+ 1)[2(n+ 1)- 1][(n+ 1)- 1]= n(n+ 1)(2n+ 1)

So 1? +2? + +n？ =n(n+ 1)(2n+ 1)/6 .