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Ta có: $a_{k}=\dfrac{3k^{2}+3k+1}{(k^{2}+k)^{3}}=\dfrac{(k^{3}+3k^{2}+3k+1)-k^{3}}{[k(k+1)]^{3}}=\dfrac{(k+1)^{3}}{k^{3}(k+1)^{3}}-\dfrac{k^{3}}{k^{3}(k+1)^{3}}$
$=\dfrac{1}{k^{3}}-\dfrac{1}{(k+1)^{3}}$
$S=a_{1}+a_{2}+ ... + a_{2006}$
$=\left(\dfrac{1}{1^{3}}-\dfrac{1}{2^{3}}\right) +\left(\dfrac{1}{2^{3}}-\dfrac{1}{3^{3}}\right)c+ ... + \left(\dfrac{1}{2006^{3}}-\dfrac{1}{2007^{3}}\right)$
$=1-\dfrac{1}{2007^{3}}$
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