+0  
 
+1
209
6
avatar+1442 

Compute  1 - 2 + 3 - 4 + ... + 2005 - 2006 + 2007.

 

and

 

Compute the sum

 

                2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5)

 

 

We didn't learn these in class... thanks for all the help today!

AnonymousConfusedGuy  Jan 16, 2018
 #1
avatar+89876 
+1

First one

 

[1 - 2] + [ 3 - 4] + ... + [2005 - 2006] + 2007

 

Note  that  each  pair of terms in the brackets  sum to  -1

 

And we have   2006/2  =  1003 of these pairs....so the series sums to  :

   

1003 (-1)  +  2007    =  1004  

 

 

Second one.....

 

2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) =

 

2/6   +  2/24   +  2/60  =

 

1/3   +  1/12  +  1/30  =        

 

20/60   +  5/60  +  2/60   =

 

27  / 60 

 

 

cool cool cool

CPhill  Jan 16, 2018
 #2
avatar+1442 
+2

Thanks!  I now realize I forgot to add this to the second problem.  Suppose that the second sequence was infinitely repeating (+ … ), what would be the sum then?

AnonymousConfusedGuy  Jan 16, 2018
 #3
avatar+89876 
0

Sorry, ACG...I know that the infinite sum of the second series is  1/2, but I don't know how to prove it......heureka  may know how to prove this.....I'll  send him a message....

 

 

cool cool cool

CPhill  Jan 16, 2018
edited by CPhill  Jan 16, 2018
 #5
avatar+1442 
+2

That would be great.  Thanks for helping out!

AnonymousConfusedGuy  Jan 16, 2018
 #4
avatar
+1

∑2/[n(n+1)(n+2)] from n=1 to infinity, then it converges to 1/2.

 

Sorry, same here, except that the Partial Sum Formula would be:  (n^2 + 3n)/(2 (n + 1) (n + 2)). So, if n=1,000,000, then:[1,003,000/2,006,004] =0.49999900.............etc.

Guest Jan 16, 2018
edited by Guest  Jan 16, 2018
edited by Guest  Jan 16, 2018
 #6
avatar+20024 
0

Compute the sum    2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5)+...

 

see link: http://web2.0calc.com/questions/algebra_47009

 

In General:

\(\begin{array}{lcll} s = \dfrac{1}{1 \cdot 2 } + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4 } + \dfrac{1}{4 \cdot 5 } + \cdots \ + \dfrac{1}{n \cdot (n+1)} + \cdots =\ \frac{1}{1!0!}\cdot \frac{1}{1} = 1 \\\\ s =\mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \frac{1}{2!0!}\cdot \frac{1}{2} } = \frac{1}{4} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3)} + \cdots =\ \frac{1}{3!0!}\cdot \frac{1}{3} = \frac{1}{18} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3 \cdot (n+4) } + \cdots =\ \frac{1}{4!0!}\cdot \frac{1}{4} = \frac{1}{96} \\\\ \ldots \\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdots } + \dfrac{1}{2 \cdot 3 \cdot 4 \cdots } + \dfrac{1}{3 \cdot 4 \cdot 5 \cdots} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)\cdot~\cdots ~\cdot (n+m) } + \cdots =\ \frac{1}{m!0!}\cdot \frac{1}{m} = \frac{1}{m\cdot m!} = \frac{1}{(m+1)!-m!} \\\\ \end{array} \\ \)

 

 

 

 

laugh

heureka  Jan 19, 2018

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.