Question

$\underset{n\to \infty }{lim}\frac{2n^2{A}_n-n{B}_n-C_n}{n^6}$ equals?

• A. $\frac{1}{20}$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. $\frac{2}{15}$

Answer 1

Answer: (D)

$\frac{2}{15}$

$A_n=1^3+2^3+3^3+...+n^3=\frac{n^2{(n+1)}^2}{4}$
$B_n=1^4+2^4+3^4+...+n^4=\frac{n(n+1)(2n+1)}{6}.\frac{3.n^2+3n-1}{5}$
$C_n=1^5+2^5+3^5+...+n^5=\frac{n^2{(n+1)}^2}{4}.\frac{2.n^2+2n-1}{3}$
$\underset{n\to \infty }{lim}\frac{2n^2{A}_n-n{B}_n-C_n}{n^6}$
$=\underset{n\to \infty }{lim}\frac{1}{n^6}\{\frac{2n^4.{(n+1)}^2}{4}-\frac{n^2(n+1)(2n+1)(3n^2+3n-1)}{30}-\frac{n^2{(n+1)}^2(2n^2+2n-1)}{12}\}$
$=\underset{n\to \infty }{lim}\frac{1}{2}{\left(\frac{n+1}{n}\right)}^2-\frac{1}{30}\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)\frac{(3n^2+3n-1)}{n^2}-\frac{1}{12}{\left(\frac{n+1}{n}\right)}^2\frac{(2n^2+2n-1)}{n^2}$
$=\frac{1}{2}-\frac{6}{30}-\frac{2}{12}=\frac{2}{15}$
Hence, option 'D' is correct.