# Prove that the sum of the squares of the first n integers is 1/24 (2n) (2n+1) (2n+2)?

Relevance

now sum = 1/6(n*(n+1)* (2n+1) = 1/24(2n* (2n+2) * (2n+ 1) = 1/24 (2n) (2n+1) (2n+2)

Proved

extending it we get ((2n+2)P3)/4

• Raj K
Lv 7

∑n² = 1² +2²+3²+4²+................+n²

Using the identity

n³ −(n−1)³= n³ −(n ³−3n²+3n−1)

→n³ −(n−1)³= 3n²−3n+1

→(n−1)³ −(n−2)³= 3(n−1)²−3(n−1)+1

....................................

.................................

.......................................

→(1)³ −(0)³= 3(1)²−3(1)+1

n³ = 3∑n²−3∑n+∑1

→ 3∑n²= n³+3∑n−∑1 Since

=n³+3n(n+1)/2−n Since ∑n=n(n+1)/2 and ∑1=n

=n{n²+3(n+1)/2−1} Simplifying

=n(2n²+3n+3−1)/2 Simplifying

=n(2n²+3n+2)/2

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

or ∑n²=n(2n+1)(n+1)/6

or ∑n²=4×n(2n+1)(n+1)/(6×4) Multiply and divide by 4

=2n(2n+1)(2n+2)/(24)

=(2n)(2n+1)(2n+2)/(24)

=(1/24)(2n)(2n+1)(2n+2)/(24)

As before, we set up as follows:

sumSquares1.gif

Saying the sum to n is one term less than the sum to (n+1)

We expand the sums:

sumSquares2.gif

As expected, the cubic terms cancel, and we rearrange the formula to have the sum of the squares on the left:

sumSquares3.gif

Expanding the cube and summing the sums:

sumSquares4.gif