Question

Sum of 1/(1*2*3) + 1/(2*3*4) +....+1/(18*19*20)

Answer 1

Answer: 189/760

Explanation:

The series can be represented in sigma notation as:

`\sum^(18)_(n=1) (1)/(n(n+1)(n+2))`

We can perform partial fraction decomposition as follows:

`(1)/(n(n+1)(n+2))=(A)/(n)+(B)/(n+1)+(C)/(n+2)\\\\ \implies 1=A(n+1)(n+2)+B(n)(n+2)+C(n)(n+1)`

If n = 0, then
`1=A(0+1)(0+2) \implies A=(1)/(2)`

If n = -1, then
`1=B(-1)(-1+2) \longrightarrow B=-1`

If n = -2, then
`1=C(-2)(-2+1) \longrightarrow C=(1)/(2)`

This means the series can be expressed as:

`\sum^(k)_(n=1) \left((1)/(2n)-(1)/(n+1)+(1)/(2(n+2)) \right)=(k(k+3))/(4(k+1)(k+2))`

Substituting in k=18,

`(18(21))/(4(19)(20)=\boxed{(189)/(760)}`