Question

For the series $2+12+36+80+150+252+.$..........., the value of $\underset{n\to \infty }{lim}\frac{T_n}{n^3}=$

Answer 1

$2+12+36+80+150+252+...\dots$

as $2=1.2=1^2\times 2$

$12=4.3=2^2\times 3$

$36=9.4=3^2\times 4$

$80=16.5=4^2\times 5$

$\therefore a_n=n^2\times (n+1)$

now

$\underset{n\to \infty }{lim}\frac{a_n}{n^3}$

$=\underset{n\to \infty }{lim}\frac{n^2(n+1)}{n^3}$

$=\underset{n\to \infty }{lim}(1+\frac{1}{n})=1$.