Can You Solve These Challenging Math Club Problems?

[sabanation12]

We were given a worksheet just to see how much we knew and sucked at it, so now I am putting some time into trying to learn this stuff. Can you guys please answer these and tell me how you got them? and there is one i would love if you guys could check :)

First Question:

"Find the Sum of the following fractions"

1/2 + 1/6 + 1/12 + 1/20 + ... + 1/9900

"Express your answer as a common fraction"

Problem Two (is there a quick way you guys do this one?)

Julie begins counting backwards from 1000 by 2's and at the same time Tony begins counting foreward by 3's, if they count at the same rate what number will they say at the same time?

Problem 3 (check my work)

"A chord of the larger of two concentric circles is tangent to the smaller circle and measure 18 inches. Find the number of square inches in the area of the shaded region (area between outer rim of the inside circle and outer rim of the larger one). Express your answer in terms of ∏.

I did: 9^2 + r^2 = R^2 (r is radius of smaller circle R is radius of larger) and so then i simplified a little to get 81 = R^2 - r^2. Then ∏R^2 - ∏r^2 = ∏(R^2-r^2) and finally got ∏(81) as my final answer.

 

[Sanket Karnik]

1/2+1/6+1/12+1/20+.....+1/9900
note that the numbers 2,6,12,20,...,9900 are of the form n(n+1)
so, nth term T_n = 1/n(n+1)
= 1/n - 1/(n+1)
so, the series is 1/1-1/2+1/2-1/3+1/3-1/4+....-1/99+1/99-1/100
= 1 - 1/100
= 99/100

 

[Vardd]

Question 2 :

1000 - (2x) = 3x
Solve for x
x = 200
Test it :
1000 - (2*200) = 600
3*200 = 600

So they say 600 at the same time.

 

[sabanation12]

n(n+1)

Please elaborate, is "n" the denominator of the previous number?

 

[Vardd]

n(n+1)

Please elaborate, is "n" the denominator of the previous number?

n=1 : 1(n+1) = 2
n=2 2(2+1) = 6
n=3 3(4) = 12
n=4 4(5) = 20
...
So n simply represents the set of number going 1 to 100.
Then we can reduce the series by using the same rule as a telescopic series to get the sum.

http://en.wikipedia.org/wiki/Telescoping_series

 

[sabanation12]

So the next number in the series "1/2 + 1/6 + 1/12 + 1/20" would be 1/30? (5+1)*5

 

[Vardd]

Correct.