It looks like you're trying to evaluate the sum,
[tex]\displaystyle \frac2{4\times7} + \frac2{7\times10} + \frac2{10\times13}+\cdots+\frac2{49\times52}[/tex]
which can be written as
[tex]\displaystyle \sum_{n=1}^{16} \frac2{(3n+1)(3n+4)}[/tex]
Split up the summand into partial fractions:
[tex]\displaystyle \frac2{(3n+1)(3n+4)} = \frac a{3n+1} + \frac b{3n+4} \\\\ \implies 2 = a(3n+4)+b(3n+1) \\\\ \implies 2 = (3a+3b)n+4a+b[/tex]
so that
3a + 3b = 0, or a = -b
4a + b = 2
Solve for a and b :
4a + (-a) = 3a = 2 ==> a = 2/3 ==> b = -2/3
So the sum is
[tex]\displaystyle \frac23 \sum_{n=1}^{16} \left(\frac1{3n+1} - \frac1{3n+4}\right)[/tex]
Write out the first terms and observe that several terms cancel with each other:
2/3 (1/4 - 1/7)
+ 2/3 (1/7 - 1/10)
+ 2/3 (1/10 - 1/13)
+ …
+ 2/3 (1/43 - 1/46)
+ 2/3 (1/46 - 1/49)
+ 2/3 (1/49 - 1/52)
So the sum collapses and simplifies to
[tex]\displaystyle \sum_{n=1}^{16} \frac2{(3n+1)(3n+4)} = \frac23 \left(\frac14 - \frac1{52}\right) = \boxed{\frac2{13}}[/tex]
