Show that

(1² + 2·2² + 3·3² + … + n·(n+1)²) / (1² + 2² + 3² + … + (n+1)²) = (3n + 5) / (3n + 1)

nᵗʰ term of the numerator = n(n + 1)² = n³ + 2n² + n

nᵗʰ term of the denominator = n² + (n + 1)² = 2n² + 2n + 1

Sum of numerator terms:

Σ(n³ + 2n² + n) = [n²(n + 1)² / 4] + [n(n + 1)(2n + 1) / 3] + [n(n + 1) / 2]

= (n(n + 1)/12)[3n² + 11n + 10]

Sum of denominator terms:

Σ(2n² + 2n + 1) = [n(n + 1)(2n + 1)/3] + [n(n + 1)/2] + n

= (n(n + 1)/6)[4n + 5]

Therefore,

(1² + 2·2² + 3·3² + … + n·(n+1)²) / (1² + 2² + 3² + … + (n+1)²) = [(n(n + 1)/12)(3n² + 11n + 10)] / [(n(n + 1)/6)(4n + 5)]

= (3n² + 11n + 10) / (2(4n + 5))

= (3n + 5) / (3n + 1)

Hence proved.