A) \[\frac{{{2}^{n+1}}-1}{{{4.2}^{n}}}\]
B) \[{{2}^{n+1}}-1\]
C) \[\frac{{{2}^{n+1}}-1}{{{3.2}^{n}}}\]
D) None of these
Correct Answer: B
Solution :
[b] \[{{x}^{2}}+4x+5={{(x+2)}^{2}}+1\ge 1.\] so, \[a=1\] \[b=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}\theta }{{{\theta }^{2}}}=2\] \[\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}={{b}^{n}}+a{{b}^{n-1}}}+{{a}^{2}}{{b}^{n-2}}+...+{{a}^{n}}\] \[=\frac{{{b}^{n}}\left[ 1-{{\left( \frac{a}{b} \right)}^{n+1}} \right]}{1-\frac{a}{b}}=\frac{{{2}^{n}}\left[ 1-{{\left( \frac{1}{2} \right)}^{n+1}} \right]}{1-\frac{1}{2}}\] \[=\frac{{{2}^{n+1}}({{2}^{n+1}}-1)}{{{2}^{n+1}}}=({{2}^{n+1}}-1).\]
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