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Home  >>  EAMCET  >>  Mathematics
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For $|x| < 1$, the constant term in the expansion of $\large\frac{1}{(x+1)^2(x-2)}$ is :

\[\begin {array} {1 1} (1)\;2 & \quad (2)\;1 \\ (3)\;0 & \quad (4)\;-\frac{1}{2} \end {array}\]
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(4) $-\large\frac{1}{2}$
answered Nov 7, 2013 by pady_1
 
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