If \[a=\min \{{{x}^{2}}+4x+5,x\in R\}\]and \[b=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{1-\cos 2\theta }{{{\theta }^{2}}},\] then the value of \[\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}\] is

\[\frac{{{2}^{n+1}}-1}{{{3.2}^{n}}}\]

\[\frac{{{2}^{n+1}}-1}{{{4.2}^{n}}}\]

If \[a=\min \{{{x}^{2}}+4x+5,x\in R\}\]and \[b=\underset{\theta \to 0}

1.	If \[a=\min \{{{x}^{2}}+4x+5,x\in R\}\]and \[b=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{1-\cos 2\theta }{{{\theta }^{2}}},\] then the value of \[\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}\] is
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
Answer» C. \[\frac{{{2}^{n+1}}-1}{{{3.2}^{n}}}\]

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If \[a=\min \{{{x}^{2}}+4x+5,x\in R\}\]and \[b=\underset{\theta \to 0}{\mathop{\lim }}\,\frac{1-\cos 2\theta }{{{\theta }^{2}}},\] then the value of \[\sum\limits_{r=0}^{n}{{{a}^{r}}.{{b}^{n-r}}}\] is

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