Answer :

__Taking L.H.S__

first we will solve the numerator & denominator separately

Let numerator be

S_{1} = 1 × 2^{2} + 2 × 3^{2} + 3 × 4^{2} + … + n × (n + 1)^{2}

n^{th} term is n × (n + 1)^{2}

Let a_{n} = n(n + 1)^{2}

= n(n^{2} + 2n + 1)

= n^{3} + 2n^{2} + n

Now, S_{1}

Let denominator be

S_{2} = 1^{2} × 2 + 2^{2} × 3 + … + n^{2} × (n + 1)

n^{th} term is n^{2} × (n + 1)

Let b_{n} = n^{2}(n + 1) = n^{3} + n^{2}

Now,

S_{2}

Now,

L.H.S

= R.H.S

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

Hence Proved.

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