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If 100∑n=1tan−1(n2+n+1)=aπb−tan−1c, where a,b,c∈N and a and b are relatively prime, then the value of a−cb is |
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Answer» If 100∑n=1tan−1(n2+n+1)=aπb−tan−1c, where a,b,c∈N and a and b are relatively prime, then the value of a−cb is |
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