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Q. The absolute term in the quadratic expression
$ \sum_{k=1}^n\left(x-\frac{1}{3 k+1}\right)\left(x-\frac{1}{3 k-2}\right) \text { as } n \rightarrow \infty \text { is } $

Complex Numbers and Quadratic Equations

Solution:

$ f (0)+ f (-1)= c + c -5=2 c -5$
Now $2 c$ is always even so $2 c -5$ is odd. may or not be divisible by 3