802 Sums of consecutive squares
29.1 Sum of squares of natural numbers ...........
Theorem 29.1.
12 +2^2 +3^2 +···+n^2 =
1
6
n(n+ 1)(2n+1).
Proof.LetTn=1+2+3···+n=^12 n(n+1)and
Sn=1^2 +2^2 +3^2 +···+n^2.
23 =1^3 +3· 12 +3·1+1
33 =2^3 +3· 22 +3·2+1
43 =3^3 +3·^32 +3·3+1
..
.
n^3 =(n−1)^3 +3(n−1)^2 +3(n−1) + 1
(n+1)^3 = n^3 +3·n^2 +3·n +1
Combining these equations, we have
(n+1)^3 =1^3 +3Sn+3Tn+n.
SinceTnis known, we have
Sn=
1
3
(
(n+1)^3 −n− 1 −
3
2
n(n+1)
)
=
1
6
n(n+ 1)(2n+1).
Exercise
1.Find 12 +3^2 +5^2 +···+(2n−1)^2.
2.Findnso thatn^2 +(n+1)^2 is a square.
