Q26 of 97 Page 199

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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