Compute

\(\mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}}}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\ \begin{array}{|lcll|} \hline s_n = \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{3}{2 \cdot 3 \cdot 4} + \dfrac{5}{3 \cdot 4 \cdot 5} + \dfrac{7}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

We rearrange:
\(\begin{array}{|rcll|} \hline \dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} &=&\left(\dfrac{2n - 1}{n}\right)\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \left(\dfrac{2n - 1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \left(2-\dfrac{1}{n}\right)\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{n(n+1)}+\dfrac{1}{n(n+2)} \\\\ &=& \dfrac{2}{n+1}-\dfrac{2}{n+2}-\left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)+ \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{n}+\dfrac{1}{2}\cdot \dfrac{1}{n}+\dfrac{2}{n+1}+\dfrac{1}{n+1}-\dfrac{2}{n+2}-\dfrac{1}{2}\cdot \left(\dfrac{1}{n+2}\right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-\dfrac{5}{2(n+2)} \quad | \quad \dfrac{1}{2(n+2)} = \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1}-5\cdot \dfrac{1}{n}\cdot \left( \dfrac12 - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{1}{2n} + \dfrac{3}{n+1} -\dfrac{5}{2n} + \dfrac{5}{n(n+2)} \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{n(n+2)} \quad | \quad \dfrac{1}{n(n+2)} = \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + 5\cdot \dfrac{1}{2}\cdot \left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) \\\\ &=& -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} \\\\ \mathbf{\dfrac{2n - 1}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ -\dfrac{3}{n} + \dfrac{3}{n+1} + \dfrac{5}{2n} - \dfrac{5}{2(n+2)} } \\ \hline \end{array}\)

Telescoping series:

\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{-3} &\color{red}+& \color{red}\dfrac{3}{2} &+& \mathbf{\dfrac{5}{2}\cdot \dfrac{1}{1}} &\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{3} \\\\ &\color{red}-& \color{red}\dfrac{3}{2} &\color{red}+& \color{red}\dfrac{3}{3} &+&\mathbf{\dfrac{5}{2}\cdot \dfrac{1}{2}}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{4} \\\\ &\color{red}-& \color{red}\dfrac{3}{3} &\color{red}+& \color{red}\dfrac{3}{4} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{3}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{5} \\\\ &\color{red}-& \color{red}\dfrac{3}{4} &\color{red}+& \color{red}\dfrac{3}{5} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{4}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{6} \\\\ &\color{red}-& \color{red}\dfrac{3}{5} &\color{red}+& \color{red}\dfrac{3}{6} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{5}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{7} \\\\ &\color{red}-& \color{red}\dfrac{3}{6} &\color{red}+& \color{red}\dfrac{3}{7} &\color{green}+&\color{green}\dfrac{5}{2}\cdot \dfrac{1}{6}&\color{green}-& \color{green}\dfrac{5}{2}\cdot \dfrac{1}{8} \\\\ && \ldots \\\\ &\color{red}-& \color{red}\dfrac{3}{n-3} &\color{red}+& \color{red}\dfrac{3}{n-2} &\color{green}+&\color{green}\dfrac{5}{2(n-3)}&\color{green}-& \color{green}\dfrac{5}{2(n-1)} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-2} &\color{red}+& \color{red}\dfrac{3}{n-1} &\color{green}+&\color{green}\dfrac{5}{2(n-2)}&\color{green}-& \color{green}\dfrac{5}{2n} \\\\ &\color{red}-& \color{red}\dfrac{3}{n-1} &\color{red}+& \color{red}\dfrac{3}{n} &\color{green}+&\color{green}\dfrac{5}{2(n-1)}&-& \mathbf{\dfrac{5}{2(n+1)}} \\\\ &\color{red}-& \color{red}\dfrac{3}{n} &+& \mathbf{\dfrac{3}{n+1}} &\color{green}+&\color{green}\dfrac{5}{2n}&-& \mathbf{\dfrac{5}{2(n+2)}} \\ \hline \end{array}\)

The colored terms shorten out

So \(s_n\) is, we have all black terms left :
\(\begin{array}{|rcll|} \hline s_n &=& -3 + \dfrac{5}{2} + \dfrac{5}{2}\cdot \dfrac{1}{2} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{3}{4} - \dfrac{5}{2(n+1)} + \dfrac{3}{n+1} -\dfrac{5}{2(n+2)} } \\ \hline \end{array} \)

\( \lim \limits_{n\to \infty} { \dfrac{5}{2(n+1)}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{3}{n+1} } = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{5}{2(n+2)} } = 0\)

\(\begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{3}{4} - 0 + 0 -0 \\\\ &=& \dfrac{3}{4} \\ \hline \end{array}\)

\(\begin{array}{lcll} \mathbf{\huge{\sum \limits_{n = 1}^{\infty} \dfrac{2n - 1}{n(n + 1)(n + 2)}} = \dfrac{3}{4} } \\ \end{array}\\ \)

"I can't figure out how to solve for ∫ sqrt(2x^2-4) / (7x). Can someone please help me?"

Like so:

(As it is an indefinite integral you should add an arbitrary constant of course.)

.

See the following image :

Note that  XY is parallel to ED, YZ is parallel to DC and XZ is parallel to EC .... therefore triangle  XYZ is similar to triangle EDC

And since X,Y  bisect FE and FD in triangle EFD  and XY is parallel to ED then

Triangle FXY is similar to triangle FDE...and....

XY / FY  =  ED / 2FY

XY = ED / 2

2XY = ED

And since XYZ is similar to EDC.....the altitude of EDC = 2*altitude of XYZ

And since XY, ED and IH   are parallel.....the altitude of triangle EDC will be the same altitude as in triangle IHC

So  the altitude of triangle CXY =

altitude of triangle XYZ + altitude of triangle EDC =

altitude of triangle XYZ +  2* altitude of triangle XYZ  =

3*altitude of triangle XYZ

And triangle CXY is on the same base as triangle XYZ....so their areas are to each other as their altitudes

So [ CYX ] = 3 * [ XYZ ] =  3(21)  =  63

What is the minimum value of the expression

 \(\mathbf{\huge{2x^2+3y^2+8x-24y+62}}\) 

for real x and y?

\(\begin{array}{|rcll|} \hline \mathbf{f(x,y)} & \mathbf{=} & \mathbf{2x^2+3y^2+8x-24y+62} \\\\ f_x=\dfrac{\partial f} {\partial x} &=& 4x+8 \\ f_y=\dfrac{\partial f} {\partial y} &=& 6y-24 \\ f_{xx}= \dfrac{\partial^2 f} {\partial x^2} &=& 4 \\ f_{yy}= \dfrac{\partial^2 f} {\partial y^2} &=& 6 \\ f_{xy}=f_{yx}= \dfrac{\partial^2 f} {\partial x\partial y}= \dfrac{\partial^2 f} {\partial y\partial x} &=& 0 \\ \hline \end{array} \)

if \(f_{xx}f_{yy}-f^2_{xy}>0\)  either a maximum or a minimum.

Distinguish between these as follows:

if \(f_{xx}<0\) and \(f_{yy} <0\)  then is a maximum point.

if \(f_{xx}>0\) and \(f_{yy} >0\)  then is a minimum point.

\(\begin{array}{|rcll|} \hline && f_{xx}f_{yy}-f^2_{xy} \\ &=& 4\cdot 6 - 0^2 \\ &=& 24 \quad | \quad >0 \quad \text{either a maximum or a minimum} \\\\ && f_{xx} \\ &=& 4 \\ && f_{yy} \\ &=& 6 \quad | \quad f_{xx}>0 \text{ and } f_{yy} >0 \quad \text{it is a minimum point}\\ \hline \end{array} \)

The minimum value:

\(\begin{array}{|rcll|} \hline f_x(x,y) = 4x+8 &=& 0 \\ 4x+8 &=& 0 \\ 4x &=& -8 \\ x &=& -\dfrac{8}{4} \\ \mathbf{x_{\text{minimum}}} &\mathbf{=}& \mathbf{ -2 } \\\\ f_y(x,y) = 6y-24 &=& 0 \\ 6y-24 &=& 0 \\ 6y &=& 24 \\ y &=& \dfrac{24}{6} \\ \mathbf{y_{\text{minimum}}} &\mathbf{=}& \mathbf{ 4 } \\ \hline \end{array} \)

2x^2-4>0

x^2-2>0

x^2>2

x>sqrt2    or   x<-sqrt2

I have not worked it out but I am thinking that you could let    \(\theta=sec^{-1}\frac{x}{\sqrt2}\)

that should help.    

The minumum value of   a    will be at   -b/2a    (this 'a' is the coefficient of a^2....1)

min val ue of a   =   -6/2 = -3   

sub this value of a into the equation to find the minumunm value of the function    f(a) =  -3^2 + 6(-3) - 7 = -16 is the min value of the fxn

This is an upward opening parabola (like a bowl.) because the x^2 coefficient is positive

The minimum will occur at x = -b/2a

- (-6)/2 = 3     use this value to find f(x)        3^2 - 6(3) + 13 = 9-18+13 = 4 is the minimum value of the expression

I believe this describes an 'ordinary' annuity with payments at the end of the month

months = 50 yrs x 12 = 600 months

monthly interest = .054 / 12 = .0045 % /mo

FV = 400,000

FV = P [ ((1+i)^n-1)/i )      400000 = P (1.0045)^600 -1)/.0045       results in P = $130.53 / mo     

y = x^2 + 12x + 5

In the form    ax^2 + bx + c.....the x value that minimizes y is given by   -b / (2a)

So.....the x value that minimizes y is given by   -12 / (2 * 1)  =  -12 / 2  = -6

Put this into the function and we get that y =   (-6)^2 + 12(-6) + 5 =   36 - 72 + 5 =  -36 + 5  = -31

k = (6x + 12)(x - 8)     expand this

k = 6x^2 + 12x - 48x - 96

k = 6x^2 - 36x - 96

The minimum value for k occurs at   x = - [ -36] / [ 2 * 6 ]  = 36 /12  = 3

So....the minimum value of k is 

6(3)^2 - 36(3) - 96 =  

54 - 108 -  96 =

-54 - 96 =

-150

The radius of the wheel is  41 + 3  = 44cm

Each rotation the wheel travels  2pi (44) =   88pi cm

So....the number of rotations is

96712  / [88pi ]  ≈  350  rotations

\(\text{3 oranges or 6 apples are mutually exclusive events so}\\ P[\text{3 oranges} \cup \text{6 apples}] = P[\text{3 oranges}]+P[\text{6 apples}]\)

\(\text{There are a total of }n=\dbinom{8+3-1}{3-1}=45 \text{ ways of choosing the 8 fruits}\\ \text{Select 3 oranges. We now have 5 pieces of fruit to choose from either banana or apple}\\ \text{there are }\dbinom{5+2-1}{2-1}=6 \text{ ways this can be done, with a probability of }\dfrac{6}{45}=\dfrac{2}{15} \text{Select 5 apples. We now have 3 pieces of fruit to choose from either orange or banana}\\ \text{There are }\dbinom{3+2-1}{2-1}=4\text{ ways of doing this with a probability of }\dfrac{4}{45}\\ P[\text{3 oranges}\cup \text{6 apples}] = \dfrac{2}{15}+\dfrac{4}{45} = \dfrac{2}{9}\)

pr = 12*sin60 = 12 * [sqrt(3)/2]  =  6 * sqrt(3)

And we can find XR thusly :

XR/ sin30  = 6/ sin60

XR   = 6sin30/ sin60

XR  = 6(1/2) / [sqrt(3) / 2]

XR  = 3*2/ sqrt(3) = 6/sqrt(3)

So PX  =  PR - XR =   6*sqrt(3) - 6/sqrt(3) =  [18 - 6] / sqrt(3)  =  12/sqrt(3)

And the area of PQX   =  (1/2) *PX* RQ  = (1/2) (12/sqrt(3)) * (6)   =  36 / sqrt(3)  square units

https://web2.0calc.com/questions/in-pqr-we-have-p-30-q-60-r-90-point-x-is-on-pr-such-that-qx-bisects-pqr-if-pq-equals-12-then-what-is-the-area-of-pqx

√72x^9y^4  =  √2 * √36 *x^9 * y^4

We can write

√2* (36)^(1/2) * (x^9)^(1/2) * (y^4)^(1/2 )  =

√2 *  6   * x^(9/2) * (y)^(2)

The only thing here is to find out how many x's stay in and how many come out

9/2 = 4 + 1/2        4 powers come out and 1 power stays in......so we have

6x^4y^2 √ [ 2 x ]

https://web2.0calc.com/questions/help_78906 

THERE !!!

trying to

No prob....!!!

In this case, we are given sample size n and probability p so this is a binomial distribution.

We can use the normal approximation where the mean is np and the standard deviation is the square root of npq where q=1-p. Then mean is 150*.09=13.5 and the standard deviation is the square root of 12.285 or 3.5.

I used Excel here but basically, I calculated the binomial probability with n=150, p=.09, and x=15 to be: .099

The cumulative binomial probability with n=150, p=.09, and x=10 is .1987

The cumulative binomial probability of having at most 25 people have electricity (n=150,p=.09, x=25) is 99.9%. So the probability of having more than 25 people is 1-.999=.0001.

Hope that helps!  

Let the bottom base be the longer side

The non-parallel side and the height form a 3-4-5 right triangle

So....using symmetry....the bottom base is (4 + 4) =  8 units longer than the top base

So.....we have

x +( x + 8 )+ 5 + 5 = 42    simplify

2x +  18 = 42

2x = 24

x = 12  =  the top base

x + 8 =  20 = the bottom base

So  

a. 12 inches and 20 inches

de nada!    Hope you are catching on !!!!!   Keep up the good work....  

Happy Birthday, Guru !  

yeah right haha