A) \[0\]
B) \[1\]
C) \[2\]
D) Does not exist
Correct Answer: B
Solution :
Here, \[0+2+4+6+....+2n\] \[=\frac{n+1}{2}[2\times 0+(n+1-1)(2)]\] \[=\frac{n+1}{2}[2n]=n(n+1)\] and \[1+3+5+7+....+(2n-1)\] \[=\frac{n}{2}[2\times 1+(n-1)(2)]\] \[=\frac{n}{2}[2+2n-2]=\frac{n}{2}\times 2n={{n}^{2}}\] Then, \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{0+2+4+6+...+2n}{1+3+5+7+...+(2n-1)}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\frac{n(n+1)}{{{n}^{2}}}=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\frac{(n+1)}{n}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\,\,\left[ 1+\frac{1}{n} \right]=1+0=1\]
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