Answer: 189/760
Step-by-step explanation:
The series can be represented in sigma notation as:
[tex]\sum^{18}_{n=1} \frac{1}{n(n+1)(n+2)}[/tex]
We can perform partial fraction decomposition as follows:
[tex]\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}\\\\ \implies 1=A(n+1)(n+2)+B(n)(n+2)+C(n)(n+1)[/tex]
If n = 0, then [tex]1=A(0+1)(0+2) \implies A=\frac{1}{2}[/tex]
If n = -1, then [tex]1=B(-1)(-1+2) \longrightarrow B=-1[/tex]
If n = -2, then [tex]1=C(-2)(-2+1) \longrightarrow C=\frac{1}{2}[/tex]
This means the series can be expressed as:
[tex]\sum^{k}_{n=1} \left(\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \right)=\frac{k(k+3)}{4(k+1)(k+2)}[/tex]
Substituting in k=18,
[tex]\frac{18(21)}{4(19)(20}=\boxed{\frac{189}{760}}[/tex]
