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10!  =  3628800

11!  = 39916800

\(S\quad{=\quad}\dfrac{6}{3^2-1}+\dfrac{6}{5^2-1}+\dfrac{6}{7^2-1}+\dfrac{6}{9^2-1}+\cdots\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{(2n+1)^2-1}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n^2+4n}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac{6}{4n(n+1)}\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n(n+1)}\bigg)\\~\\~\\ S_m\quad{=\quad}\sum\limits_{n=1}^{m}\ \dfrac32\bigg(\dfrac{1}{n}-\dfrac{1}{n+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}- \dfrac{1}{4}+\dots+\dfrac{1}{m-1}-\dfrac{1}{m}+\dfrac{1}{m}-\dfrac{1}{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(\dfrac11-\dfrac1{m+1}\bigg)\\~\\~\\ S_m\quad{=\quad}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}S_m\\~\\~\\ S\quad=\quad\lim\limits_{m\rightarrow\infty}\dfrac32\bigg(1-\dfrac1{m+1}\bigg)\\~\\~\\ S\quad=\quad\dfrac32\bigg(1-0\bigg)\\~\\~\\ S\quad=\quad\dfrac32\)_

6  [     1 / [ 3^2 - 1 ]   +   1 / [ 5^2 - 1]   +  1/[7^2  - 1] +   1/ [ 9^2 - 1 ] +....... ]   =

6 [     1 /  8    +   1 / 24    +     1/48  +  1/80  +   ..... ]    =

6  [   1 / [ 2*4]   +  1/ [ 4*6]  +  1/[ 6*8] + 1/ [8*10] +  ....... ]   =

6   [ (1/4) ]  [   1 / [1 * 2]   +  1/ [2 *3]   + 1/ [ 3*4]  + 1 / [ 4*5] +   .....]

Note that            1/ [ n * (n + 1) ]   =         1 / n   -  1/[ n + 1]

So we can write

(6/4)   [   ( 1/1 - 1/2) + (1/2 - 1/3)  + (1/3 - 1/4) + (1/4 - 1/5)  + .......] =

(3/2)  [   1   + ( 1/2 -1/2) + (1/3 - 1/3)   + (1/4 - 1/4)  + ........ +  (-1/n) ]  = 

(3/2) [ 1  -   1/n ]         as n ⇒  infinity,   1/n   ⇒   0

So we have

(3/2 )  [ 1 - 0]  =

(3/2) (1)   =

3/2

sumfor(n, 1, 1000000, 6 / ((2*n + 1)^2 - 1)) = 3 / 2

As n grows toward infinity, the 1/n+1 vanishes to 0, so the sum of the infinite series is 3/2

Wait, I can label the bottom parts as 1/2a and 1/2a, and I can label the top parts as 1/2b and 1/2b.

There are also two triangles at the top, so I can use the Pythagorean Theorem? 

Consider...if the side of the square is 7.3 m....the area   = 7.3^2  = 53.29 m^2

So......this is equal to the area of one of the circular tables

So.....

53.29  = pi * r^2

53.29/pi =  r^2       take the square root of both sides

r ≈  4.1 m

I'm wondering if this  was meant in the original problem instead of  a circumference of 4.1  m  ???

I hate to be the one to tell the textbook people they're wrong, but they're wrong.

I worked it myself, and CPhill is right on the money.  The side of each table is 1.16 meter.

Just think about it.  Every table is 7.3 meters square?  Every table is 24 FEET wide?  I don't think so. 

Check and see if you accidentally looked at the wrong answer in the textbook.

.

Wow CPHILL.. Great job!

Wow, thanks for responding so quickly but unfortunately the answer is 7.3m instead of 1.16m and I'm not sure why it is, the text book only reveals the answer instead of a methodical guide. 

Circumference  =  2 pi * r

4.1  =  2 pi * r       divide both sides by 2pi

4.1 / [ 2pi]  = r

So....the area of one table is just the area of a circle  =  pi  r^2  =   pi *  [  4.1 / [2pi]  ]^2  =

 [ 4.1 / 2] ^2 / pi    m^2

And this is equal to the area of the square

So....the side of this square is just the square root of this  =  [ 4.1/2] / pi^(1/2) ≈  1.16 m

4) With a tailwind, a small plane traveled 420 miles in one hour and 30 min. The return trip took 30 min longer. Find the wind speed and airspeed of the plane...

This one is similar to (3)

1hr 30 min  = 1.5 hrs.....so

(R + W) * 1.5  =  420     (1)

(R-W )  *2  =   420    ⇒ 2R - 2W  = 420   ⇒  R - W  = 210  ⇒  R  = 210 + W   (2)

Sub (2) into (1)   for R  and we have that

(210 + W + W) * 1.5   = 420

(210 + 2W) * 1.5  =  420

315 + 3W  =  420     subtract   315 from each side

3W  =  105        divide both sides by 3

W  = 35   mph

And using (1) ,  R  =  210 + 35  =  245 mph

3) With a headwind, a plane traveled 840 miles northward in 2 hours. With the same wind as a tail wind, the return trip took 1 hour and 45 mineutes. Find the plane's airspeed and the wind speed.

Note....1hr 45 min  =  1.75 hrs

Let the speed of the plane in still air be S

Let the wind speed  be W

Against the wind the plane's rate is  (R - W)  ....the wind slows the plane's normal speed

With the wind the plane's rate is (R + W).....the wind increases the plane's normal speed

So.....  Rate * Time  =  Distance   and we have that

(R -W) * 2  =  840   ⇒  2R - 2W  =  840  ⇒   R - W  = 420  ⇒   R  = 420 + W    (1)

(R + W) *1.75  = 840  ⇒  1.75 R + 1.75W  = 840    (2)

Sub (1)  into (2)   for R  and we have that

1.75(420 + W) + 1.75W  = 840

735 + 1.75W + 1.75W  = 840

735 + 3.5W  = 840       subtract 735 from both sides

3.5W  =  105       divide both sides by  3.5

W  = 30  mph = wind speed

And using (1) the plane's speed  is  420 + 30  =  450 mph

No Problem ; )

THX, HiylinLink.......!!!

You can see part of the question here answered by Cphill: https://web2.0calc.com/questions/double-trouble-2-0

same... I was helping an user at brainly.com, then the user popped a random question at me.... I was so confused so I posted here...

Perhaps the following will make it clearer:

EP made the trivial mistake of setting 1 hour 45 mins (i.e. one and three quarter hours) as 1.45 hours rather than 1.75 hours

Now see how you get on with your question 4) . 

EP: How did you get: 

(W+420+W)*1.45 = 840

W= 30 mph ???

1 hour and 45 minutes =1 hour + 45/60 =1.75 -An hour and three-quarters. Why did you multiply by 1.45 ???.

Your W works out =2310 /29 =~79.65 ???