Show that

Taking L.H.S

first we will solve the numerator & denominator separately

Let numerator be

S₁ = 1 × 2² + 2 × 3² + 3 × 4² + … + n × (n + 1)²

n^th term is n × (n + 1)²

Let a_n = n(n + 1)²

= n(n² + 2n + 1)

= n³ + 2n² + n

Now, S₁

Let denominator be

S₂ = 1² × 2 + 2² × 3 + … + n² × (n + 1)

n^th term is n² × (n + 1)

Let b_n = n²(n + 1) = n³ + n²

Now,

S₂

Now,

L.H.S

= R.H.S

Hence, L.H.S = R.H.S

Hence Proved.