Sequencing Genome, I mean Limits

Calculus Level 3

Let $a = \min (x^2+2x+3)$ and $b = \displaystyle \lim_{x \to 0} \frac { \sin x \ \cos x}{e^x - e^{-x} }$ for real parameter $x$ .

What is the value of $\displaystyle \sum_{r=0}^n a^r b^{n-r}$ ?

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\frac{4^{n+1}-1}{3 \cdot .2^{n}}

\frac{2^{n}-1}{3 \cdot 2^{n}}

\frac{2^{n+1}+1}{3 \cdot 2^{n}}

\frac{2^{n+1}-1}{3 \cdot 2^{n}}

1 solution

Harshvardhan Mehta
Mar 16, 2015

$a=(x+1)^{2}+2 \Rightarrow a = 2$ $b=\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}\frac { sin(2x) }{ \frac { 2({ e }^{ 2x }-1).2x }{ 2x } } =\frac { 1 }{ 2 }$ $Now,\quad \sum _{ r=0 }^{ n }{ { a }^{ r }{ b }^{ n-r }=\sum { { 2 }^{ r } } } { \left( \frac { 1 }{ 2 } \right) }^{ n-r }=\frac { 1 }{ { 2 }^{ n } } \sum _{ r=0 }^{ n }{ { 2 }^{ 2r } } =\frac { 1 }{ { 2 }^{ n } } \sum _{ r=0 }^{ n }{ 4^{ r } }$ $=\frac { 1 }{ { 2 }^{ n } } \left[ 1+4+{ 4 }^{ 2 }+....+{ 4 }^{ n } \right] =\frac { 1 }{ { 2 }^{ n } } \left[ \frac { { 4 }^{ n+1 }-1 }{ 3 } \right] =\boxed { \frac { { 4 }^{ n+1 }-1 }{ 3.{ 2 }^{ n } } }$

sums of geometric progression (1+4+4^2+4^3+..+4^n) - why not (4^n-1)/3 = a(1-r^n)/(1-r). Only thinking - why is tehre 4^(n+1) in numerator?

Aneta Dóczy - 1 year, 9 months ago

Sequencing Genome, I mean Limits

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