3 years ago

Identify the recursive formula for the following sequence: 1/2, 1 1/2, 3 1/2, 7 1/2...

Mathematics

1 answer:

Virty [35]3 years ago

4 0

Answer:

Step-by-step explanation:

The problem is where does n start?

a1 = 1/2

a2 = 2*(a1) + 1/2

a2 = 2*(1/2) + 1/2

a2 = 1 + 1/2

a2 = 1 1/2

=========

a3 = 2*a2 + 1/2

a3 = 2*(1 1/2) + 1/2

a3 = 3 + 1/2

a3 = 3 1/2

==============

a4 = 2(3 1/2) + 1/2

a4 = 7 + 1/2

a4 = 7 1/2

==============

an = 2*a_(n-1) + 1/2

The answer is C

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blondinia [14]

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2 years ago

Use the cross product to find the sin(θ) where θ is the angle between v and w where v=i+j and w=i+j+k.

romanna [79]

$\|\mathbf v\times\mathbf w\|=\|\mathbf v\|\|\mathbf w\|\sin\theta$

$\mathbf v\times\mathbf w=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&0\\1&1&1\end{vmatrix}$
$\mathbf v\times\mathbf w=\begin{vmatrix}1&0\\1&1\end{vmatrix}\mathbf i-\begin{vmatrix}1&0\\1&1\end{vmatrix}\mathbf j+\begin{vmatrix}1&1\\1&1\end{vmatrix}\mathbf k$

$\mathbf v\times\mathbf w=\mathbf i-\mathbf j$
$\implies\|\mathbf v\times\mathbf w\|=\sqrt{1^2+(-1)^2}=\sqrt2$

$\mathbf v=\sqrt{1^2+1^2}=\sqrt2$
$\mathbf w=\sqrt{1^2+1^2+1^2}=\sqrt3$

$\sqrt 2=\sqrt2\cdot\sqrt3\sin\theta$
$\dfrac1{\sqrt3}=\sin\theta$