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Let [x] denote the greatest integer not exceeding the real number x , and an=(2n^2+4n+1/3n^2) , then Sigma [an] (where n=1 to 100) equals

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Solution:

$sum_n=1^100[a_n]=[frac73]+[frac1712]+[frac3127]+[frac4948]+[frac7175]+.......$

$sum_n=1^100[a_n]=2+1+1+1=5.$

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Deependra Verma

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