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