Observe that the [tex]n[/tex]-th term in the sum is
[tex]\dfrac{n^2 + n(n+1) + (n+1)^2}{n^3(n+1)^3} = \dfrac{3n^2 + 3n + 1}{n^3 (n+1)^3} \\\\ ~~~~~~~~ = \dfrac{(n+1)^3 - n^3}{n^3(n+1)^3} \\\\ ~~~~~~~~ = \dfrac1{n^3} - \dfrac1{(n+1)^3}[/tex]
Then the sum telescopes, and we have
[tex]\displaystyle \sum_{n=1}^{9} \frac{n^2 + n(n+1) + (n+1)^2}{n^3 (n+1)^3} = \sum_{n=1}^{9} \left(\frac1{n^3} - \frac1{(n+1)^3}\right) \\\\ ~~~~~~~~ = \left(\frac1{1^3} - \frac1{2^3}\right) + \left(\frac1{2^3} - \frac1{3^3}\right) + \cdots + \left(\frac1{8^3} - \frac1{9^3}\right) + \left(\frac1{9^3} - \frac1{10^3}\right) \\\\ ~~~~~~~~ = \frac1{1^3} - \frac1{10^3} \\\\ ~~~~~~~~ = 1 - \frac1{1000} = \boxed{\frac{999}{1000}}[/tex]
The sum of the series is 999/1000 .
A series is a sequence of expression in a certain pattern.
The series of n terms is given
[tex]\rm \frac{1^2+1*2+2^2}{1^3*2^3} + \frac{2^2+2*3+3^2}{2^3*3^3} +...+\frac{9^2+9*10*10^2}{9^3*10^3}[/tex]
nth term of the series is given by
[tex]\rm T_n = \rm \dfrac{n^2 + n(n+1)+ (n+1)^2}{n^3*(n+1)^3}[/tex]
On simplification it can be written as
[tex]\rm T_n = \rm \dfrac{3n^2 + 3n+1}{n^3*(n+1)^3}\\\\T_n = \rm \dfrac{3n(n +1)+1}{n^3*(n+1)^3}\\\\\\T_n = \rm \dfrac{ (n +1)^3 - n^3}{n^3*(n+1)^3}\\\\T_n = \rm \dfrac{ 1}{n^3} - \dfrac{1}{(n+1)^3}[/tex]
The sum of terms from 1 to 9 is given by
∑ ( [tex]\rm \frac{ 1}{n^3} - \frac{1}{(n+1)^3}[/tex])
= [tex]\rm \dfrac{1}{1^3} - \dfrac{1}{2^3} + \dfrac{1}{2^3} - \dfrac{1}{3^3}+ .......... + \dfrac{1}{9^3} - \dfrac{1}{10^3}[/tex]
= (1/1³) - (1/10³)
= 999/1000
Therefore the sum of the series given is 999/1000
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