# find the value of limn->infinity ((12+22+....+n2)(13+23+....n3)(14+24+.....n4))/((15+25+.....n5))2

Samyak Jain
333 Points
5 years ago
Ans. is 3/5.
Sum of squares of first n natural numbers is n(n + 1)(2n + 1) / 6,
sum of cubes of first n natural numbers is n2(n + 1)2 / 4,
that of fourth powers is (3n2 + 3n – 1).n(n + 1)(2n + 1) / 30 and
that of fifth powers is n2(n + 1)2(2n2 + 2n – 1) / 12.
In fact you can calculate sum of kth powers of first n natural numbers if you know the sum till (k–1)th powers.
$\therefore$ given limit is
limn$\dpi{80} \rightarrow$$\dpi{80} \infty$ {n(n + 1)(2n + 1)/6}{n2(n + 1)2 / 4}{(3n2 + 3n – 1).n(n + 1)(2n + 1) / 30} / {n2(n + 1)2(2n2 + 2n – 1) / 12}2
Simplify above expression and get
limn$\dpi{80} \rightarrow$$\dpi{80} \infty$ (2n + 1)2 (3n2 + 3n – 1) / 5(2n2 + 2n – 1)2
= (1/5) limn$\dpi{80} \rightarrow$$\dpi{80} \infty$ (4n2 + 4n + 1) (3n2 + 3n – 1) / (2n2 + 2n – 1)2
= (1/5) limn$\dpi{80} \rightarrow$$\dpi{80} \infty$ [(4n2 + 4n + 1) / (2n2 + 2n – 1)] x [limn$\dpi{80} \rightarrow$$\dpi{80} \infty$ (3n2 + 3n – 1) / (2n2 + 2n – 1)]
= (1/5) . (4/2) . (3/2)
= 3/5.