# 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

nth term is n × (n + 1)2

Let an = n(n + 1)2

= n(n2 + 2n + 1)

= n3 + 2n2 + n

Now, S1

Let denominator be

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

nth term is n2 × (n + 1)

Let bn = n2(n + 1) = n3 + n2

Now,

S2

Now,

L.H.S

= R.H.S

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

Hence Proved.

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