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# Help CPhill and Melody, You are awesome!!! Math

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1.)

Calculate

$\left(1 - \frac12 \right) \cdot \left(1 - \frac13 \right) \cdot \left(1 - \frac14\right) \dotsm \left(1 - \frac {1}{2009} \right).$

2.)

The following list of rows of odd numbers continues in the same pattern:
1

3,5

7,9,11

13,15,17,19
Each row starts where the previous left off, and has one more term than the previous row. There is one row in which the sum of the units digits of the numbers in that row equals . How many terms are in that row?

3.)

What is the maximum number of consecutive positive integers that can be added together to create a sum less than 400?

4.)

What is the units digit of $2^{1993} + 3^{1993}$?

5.)

What is the value of the sum

$\dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}$

6.)

Compute

$1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \dots + n \cdot \frac {1}{2^n} + \dotsb.$

Jdaye  Jan 28, 2018
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#1
+85757
+2

1)

(1 - 1/2) * (1 - 1/3) * (1- 1/4) * (1 - 1/5) *...........* ( 1 - 1/2007) * (1 - 1/2008) * (1 -1/2009)  =

(1/2) (2/3) (3/4) ( 4/5)  *...........*   ( 2006/2007) (2007/2008) (2008/ 2009) =

(1) (2/2) (3/3) (4/ 4) (5/5) *.........* ( 2006/2006) (2007/ 2007) (008/ 2008) (1/2009) =

(1) (1) (1) *...... *(1/2009)  =

(1 / 2009)

CPhill  Jan 28, 2018
#2
+85757
+2

5)

$$\dfrac {1}{1\cdot 3} + \dfrac {1}{3\cdot 5} + \dfrac {1}{5\cdot 7} + \dfrac {1}{7\cdot 9} + \cdots + \dfrac {1}{199\cdot 201}$$

Note that  we can write      1 /  [  n (n + 2) ]  as     (1/2)  [ 1/n  -  1 / (n + 2)]

So  we can write

(1/2) [  (1/1  - 1/3)  + ( 1/3 - 1/5) + ( 1/5 - 1/7) + ( 1/7 - 1/9)  + .....+ (1/197 - 1/199) + (1/199 - 1/201) ]  =

(1/2) (1/1  -  1/201)  =

(1/2) ( 201 - 1 ) / 201  =

(1/2) (200) / 201  =

100 / 201

CPhill  Jan 28, 2018
edited by CPhill  Jan 28, 2018
#3
+85757
+1

3 and 4 answered here :

CPhill  Jan 28, 2018

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