1.

if sum_(n=1)^oo tan^(-1)(1/(7-5n+n^2))=pi/2 +tan^(-1)c , then c is equal to

Answer»

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`5/2`

Solution :`sum_(n=1)^OO TAN^(-1)(1/(7-5n+n^2))=sum_(n=1)^oo "tan"^(-1) ((n-2)-(n-3))/(1+(n-2)(n-3))`
`sum_(n=1)^oo[tan^(-1) (n-2)-tan^(-1)(n-3)]`
`=pi/2-tan^(-1)(-2)=pi/2+"tan"^(-1)2=(API)/b + tan^(-1)C`
a=1,b=2,c=2


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