Question

$\underset{n\to \infty }{lim}[\frac{1^2}{1-n^3}+\frac{3}{1+n^2}+\frac{5^2}{1-n^3}+\frac{7}{1+n^2}+...uptonterms]=$

• A. $-\frac{10}{3}$
• B. $\frac{10}{3}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is C $\frac{1}{3}$

$\underset{n\to \infty }{lim}[\frac{1^2}{1-n^3}+\frac{3}{1+n^2}+\frac{5^2}{1-n^3}+\frac{7}{1+n^2}+...]$

$=\underset{n\to \infty }{lim}[\frac{1^2}{1-n^3}+\frac{5^2}{1-n^3}+\frac{9^2}{1-n^3}+...upto\frac{n}{2}terms]+\underset{n\to \infty }{lim}[\frac{3}{1+n^2}+\frac{7}{1+n^2}+...upto\frac{n}{2}terms]$

$=\underset{n\to \infty }{lim}\frac{{\sum }_{k=1}^{\frac{n}{2}}{(4k-3)}^2}{1-n^3}+\underset{n\to \infty }{lim}\frac{{\sum }_{k=1}^{\frac{n}{2}}(4k-1)}{1+n^2}$

$={lim}_{n\to \infty }\frac{{\sum }_{k=1}^{\frac{n}{2}}{(16k}^2-24k+9)}{1-n^3}+{lim}_{n\to \infty }\frac{{\sum }_{k=1}^{\frac{n}{2}}(4k-1)}{1+n^2}$

$=\underset{n\to \infty }{lim}\frac{\frac{4n(n+2)(n+1)}{6}-\frac{6n(n+2)}{2}+\frac{9n}{2}}{n^3(\frac{1}{n^3}-1)}+\underset{n\to \infty }{lim}\frac{\frac{n(n+2)}{2}-\frac{n}{2}}{n^2(1+\frac{1}{n^2})}$

$=\frac{-2}{3}+1=\frac{1}{3}$

Hence, option 'C' is correct.