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Compute \(\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{100^3 - 100}.\)

 May 18, 2020
 #1
avatar+111321 
+1

Note  that  we  can  write

 

       1                            1

__________   =      _______________ 

n ( n^2 - 1)               n ( n + 1) (n - 1)

 

 

Using  partial fractions

 

    1                                  A           B              C

___________    =         ___   +  _____  +   _____    multiply through by n(n + 1)(n -1)

n(n + 1) (n-1)                  n          n + 1         n - 1

 

 

1  =  A( n + 1) ( n - 1)    +  Bn(n - 1)   +  Cn ( n + 1)

 

1  = A( n^2 - 1)    +  B( n^2 - n)    +   C(n^2 + n)

 

1 =   (A + B + C)n^2  + (B - C)n  - A

 

Equating coefficients

 

A = -1

 

B - C  =   0    

 

A + B + C  =   0

 

Add  the second and third equations

 

A  + 2B  = 0

B = 1/2

C = 1/2

 

So  we  have

 

( -1/n  +  1/[2(n + 1)]  + 1'/[ 2 ( n - 1) ]  =

 

Sum  n   = 2   to  n = 100

 

(-1/2  +  1/ 6  + 1/2)   +

(-1/3  +  1/8   + 1/4)   +

(-1/4  +  1/10  + 1/6)  +

(-1/5  +  1/12  + 1/8) +

(-1/6  +  1/14  + 1/10) +

(-1/7  +  1/16  + 1/12

(-1/8  +  1/18   + 1/14) +  ........+

(-1/97  + 1/196  + 1/192)  +

(-1/98  + 1/198  + 1/194)  +

(-1/99 +  1/200 +  1/196)  +

( -1/100 + 1/202  + 1/198)   =

 

All the terms in red will  " cancel"  and we are left with

 

(1/4 + 1/200 + 1/202 - 1/100)

 

(50/ 200  + 1/200 + 1/202  - 2/200)  =

 

(49/200 + 1/202)  =

 

[ 49 (202 )  + 200 ]            10098            5049

_______________  =     _________ = _______

   200 *  202                      40400           20200

 

 

cool cool cool

 May 18, 2020
 #2
avatar+24949 
+1

Compute


\(\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}\).

 

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &=& \dfrac{1}{2(2^2 - 1)} + \dfrac{1}{3(3^2 - 1)} + \dfrac{1}{4(4^2 - 1)} + \dots + \dfrac{1}{100(100^2 - 1)} \\\\ &=& \dfrac{1}{2(2 - 1)(2+1)} + \dfrac{1}{3(3 - 1)(3+1)} + \dfrac{1}{4(4 - 1)(4+1)} + \dots + \dfrac{1}{100(100 - 1)(100+1)} \\\\ &=& \dfrac{1}{\mathbf{1}*2*3} + \dfrac{1}{\mathbf{2}*3*4} + \dfrac{1}{\mathbf{3}*4*5} + \dots + \dfrac{1}{\mathbf{99}*100*101} \\\\ &=& \sum \limits_{n=1}^{99}\dfrac{1}{n(n+1)(n+2)} \\\\ &=& \sum \limits_{n=1}^{99}\dfrac{1}{n(n+1)} * \dfrac{1}{(n+2)} \quad | \quad \dfrac{1}{n(n+1)}= \dfrac{1}{n}- \dfrac{1}{(n+1)}\\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{n}- \dfrac{1}{(n+1)}\right) \dfrac{1}{(n+2)} \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{n(n+2)}- \dfrac{1}{(n+1)(n+2)} \right) \quad | \quad \dfrac{1}{n(n+2)}=\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\Bigg(\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right)- \dfrac{1}{(n+1)(n+2)} \Bigg) \quad | \quad \dfrac{1}{(n+1)(n+2)}=\left( \dfrac{1}{(n+1)}- \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\Bigg(\dfrac{1}{2}\left( \dfrac{1}{n}- \dfrac{1}{(n+2)} \right)- \left( \dfrac{1}{(n+1)}- \dfrac{1}{(n+2)} \right) \Bigg) \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{2}*\dfrac{1}{n}- \dfrac{1}{2}*\dfrac{1}{(n+2)} - \dfrac{1}{(n+1)}+ \dfrac{1}{(n+2)} \right) \\\\ &=& \sum \limits_{n=1}^{99}\left(\dfrac{1}{2}*\dfrac{1}{n}- \dfrac{1}{(n+1)} + \dfrac{1}{2}*\dfrac{1}{(n+2)} \right) \\\\ &=& \dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n} \right) -\sum \limits_{n=1}^{99}\left(\dfrac{1}{n+1} \right) +\dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n+2} \right) \\ \\ &=& \dfrac{1}{2}\sum \limits_{n=1}^{99}\left(\dfrac{1}{n} \right) -\sum \limits_{n=2}^{100}\left(\dfrac{1}{n} \right) +\dfrac{1}{2}\sum \limits_{n=3}^{101}\left(\dfrac{1}{n} \right) \\ \\ &=& \dfrac{1}{2}\left(1+\dfrac{1}{2} \right)+ \dfrac{1}{2}\sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) -\dfrac{1}{2}-\dfrac{1}{100} -\sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) +\dfrac{1}{2}*\dfrac{1}{100}+ \dfrac{1}{2}*\dfrac{1}{101} + \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) \\ \\ &=& \left( \dfrac{1}{2} - 1 + \dfrac{1}{2} \right) \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) +\dfrac{1}{2} + \dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{100} +\dfrac{1}{200}+ \dfrac{1}{202} \\ \\ &=& \left( 1 - 1 \right) \sum \limits_{n=3}^{99}\left(\dfrac{1}{n} \right) + \dfrac{1}{4}-\dfrac{1}{100}+\dfrac{1}{200}+ \dfrac{1}{202} \\\\ &=&0 + \dfrac{50}{200}-\dfrac{2}{200}+\dfrac{1}{200}+ \dfrac{1}{202} \\\\ &=& \dfrac{49}{200}+ \dfrac{1}{202} \\\\ &=& \dfrac{49*202+200}{200*202} \\\\ &=& \dfrac{10098}{40400} \\\\ &=& \mathbf{\dfrac{5049}{20200} } \\ \hline \end{array}\)

 

laugh

 May 19, 2020

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