+0  
 
+1
492
6
avatar+1438 

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!

 Jan 16, 2018
 #1
avatar+111387 
+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

 Jan 16, 2018
 #2
avatar+1438 
+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+111387 
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

 Jan 16, 2018
edited by CPhill  Jan 16, 2018
 #5
avatar+1438 
+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.

 Jan 16, 2018
edited by Guest  Jan 16, 2018
edited by Guest  Jan 16, 2018
 #6
avatar+25225 
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

 Jan 19, 2018

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