Use induction method to prove that 1.2^2+2.3^2+3.4^2+...+r(r+1)^2= n(n+1)(3n^2+11n+10)/12

Base case (n = 1):

• left side = 1×2² = 4

• right side = 1×(1 + 1)×(3×1² + 11×1 + 10)/12 = 4

Induction hypothesis: Assume equality holds for n = k, so that

1×2² + 2×3² + 3×4² + … + k × (k + 1)² = k × (k + 1) × (3k ² + 11k + 10)/12

Induction step (n = k + 1):

1×2² + 2×3² + 3×4² + … + k × (k + 1)² + (k + 1) × (k + 2)²

= k × (k + 1) × (3k ² + 11k + 10)/12 + (k + 1) × (k + 2)²

= (k + 1)/12 × (k × (3k ² + 11k + 10) + 12 × (k + 2)²)

= (k + 1)/12 × ((3k ³ + 11k ² + 10k) + 12 × (k ² + 4k + 4))

= (k + 1)/12 × (3k ³ + 23k ² + 58k + 48)

= (k + 1)/12 × (3k ³ + 23k ² + 58k + 48)

On the right side, we want to end up with

(k + 1) × (k + 2) × (3 (k + 1) ² + 11 (k + 1) + 10)/12

which suggests that k + 2 should be factor of the cubic. Indeed, we have

3k ³ + 23k ² + 58k + 48 = (k + 2) (3k ² + 17k + 24)

and we can rewrite the remaining quadratic as

3k ² + 17k + 24 = 3 (k + 1)² + 11 (k + 1) + 10

so we would arrive at the desired conclusion.

To see how the above rewriting is possible, we want to find coefficients a, b, and c such that

3k ² + 17k + 24 = a (k + 1)² + b (k + 1) + c

Expand the right side and collect like powers of k :

3k ² + 17k + 24 = ak ² + (2a + b) k + a + b + c

==>   a = 3   and   2a + b = 17   and   a + b + c = 24

==>   a = 3, b = 11, c = 10