A) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n-1}}+1 \right]\]
B) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n-1}}-1 \right]\]
C) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n+1}}-1 \right]\]
D) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n+1}}+1 \right]\]
Correct Answer: A
Solution :
[a] \[{{t}_{3/4}}=\frac{1}{K(n-1)}\left[ \frac{1}{{{\left( \frac{{{C}_{0}}}{4} \right)}^{n-1}}}-\frac{1}{C_{0}^{n-1}} \right]\] \[=\frac{1}{K(n-1)}({{4}^{n-1}}-1)C_{0}^{n-1}\] \[\therefore {{t}_{1/2}}=\frac{1}{K(n-1)}({{2}^{n-1}}-1)\] So \[\frac{{{t}_{3/4}}}{{{t}_{1/2}}}={{2}^{n-1}}+1\]You need to login to perform this action.
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