(d) 4
2n +1Pn : 2n–1Pn = 3 : 5
⇒ (4n2 + 2n) 5 = 3 × (n3 + 3n2 + 2n)
⇒ 20n2 + 10n = 3n3 + 9n2 + 6n
⇒ 3n3 – 11n2 – 4n = 0
⇒ n(3n2 – 11n – 4) = 0
⇒ n(3n2 – 12n + n – 4) = 0
⇒ n(3n(n – 4) + 1(n – 4)) = 0
⇒ n = 0 or 4 or \(-\frac13\)
Ignoring n = 0, \(-\frac13\) as these are in admissible values, we have n = 4.