A) \[\frac{2}{7}\]
B) \[\frac{1}{12}\]
C) \[\frac{19}{52}\]
D) None of these
Correct Answer: A
Solution :
| [a] \[\underset{n\to \infty }{\mathop{Lim}}\,\underset{r=3}{\overset{n}{\mathop{II}}}\,\frac{{{r}^{3}}-8}{{{r}^{3}}+8}\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\left( \frac{{{3}^{3}}-8}{{{3}^{3}}+8} \right)\left( \frac{{{4}^{3}}-8}{{{4}^{3}}+8} \right)....\left( \frac{{{n}^{3}}-8}{{{n}^{3}}+8} \right)\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\left( \frac{3-2}{3+2}.\frac{{{3}^{2}}+4+2(3)}{{{3}^{2}}+4-2(3)} \right)\] |
| \[\left( \frac{4-2}{4+2}.\frac{{{4}^{2}}+4+2(4)}{{{4}^{2}}+4-2(4)} \right)...\left( \frac{n-2}{n+2}.\frac{{{n}^{2}}+4+2n}{{{n}^{2}}+4-2n} \right)\] |
| \[=\underset{n\to \infty }{\mathop{Lim}}\,\left( \frac{3-2}{3+2}.\frac{4-2}{4+2}.\frac{5-2}{5+2}....\frac{n-2}{n+2} \right)\] |
| \[\left( \frac{{{3}^{3}}+4+2(3)}{{{3}^{3}}+4-2(3)}.\frac{{{4}^{2}}+4+2(4)}{{{4}^{2}}+4-2(4)}...\frac{{{n}^{2}}+4+2n}{{{n}^{2}}+4-2n} \right)\] |
| \[=\left( \frac{1.2.3.4.5.6.7...}{5.6.7.8.....} \right)\left( \frac{19.28.39.52.63......}{7.12.19.28.39.52.....} \right)\] |
| \[=\frac{1.2.3.4}{7.12}=\frac{2}{7}\] |
You need to login to perform this action.
You will be redirected in
3 sec
