NCERT Solutions
Class 11 Maths
Sequences and Series

Ex.Misc.Q.26
Show that {1 × 22 + 2 × 23 + …...n × (n + 1)2} ÷ {12 × 2 + 22 × 2 + …...n2 × (n + 1)} = (3n + 5) ÷ (3n + 1)
nth term of numerator = n (n + 1)2 = n3 + 2n2 + n nth term of denominator = n2(n + 1) = n3 + n2
= n2(n + 1)2 ÷ 4 + 2{n (n + 1) (2n + 1)} ÷ 6 + n (n + 1) ÷ 2
= (n (n + 1) ÷ 2) [n (n + 1) ÷ 2 + 2(2n + 1) ÷ 3 + 1]
= (n (n + 1) ÷2) [3n2 + 3n + 8n + 4 + 6]/6
= (n (n + 1) ÷ 2) [3n2 + 11n + 10] ÷ 6
= (n (n + 1) ÷ 2) [3n2 + 6n + 5n + 10] ÷ 6
= (n (n + 1) ÷ 2) [3n (n + 2) + 5(n + 2)] ÷ 6
= [[n (n + 1) (n + 2) (3n + 5)] ÷ 12
Again,
= {n (n + 1) ÷ 2} [3n2 + 3n + 4n + 2] ÷ 6
= {n (n + 1) ÷ 2} [3n2 + 7n + 2] ÷ 6
= n (n + 1) [3n2 + 6n + n + 2] ÷ 12
= n (n + 1) [3n (n + 2) + 1(n + 2)] ÷ 12
= [n (n + 1) (n + 2) (3n + 1)] ÷ 12 …………. (3)
From equation (1), (2) and (3), we get
= [{n (n + 1) (n + 2) (3n + 5)} ÷ 12] ÷ [{n (n + 1) (n + 2) (3n + 1)} ÷ 12]
= (3n + 5) ÷ (3n + 1)
Thus, the given result is proved.