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Question Number 95048 by Tony Lin last updated on 22/May/20

a_1 =−(1/(30)) a_2 =−(1/(12)) a_3 =−(1/6) a_4 =−(7/(24)) a_5 =−(7/(15)) a_6 =−(7/(10)) a_7 =−1 find a_k

$${a}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{30}} \\ $$$${a}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$${a}_{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{6}} \\ $$$${a}_{\mathrm{4}} =−\frac{\mathrm{7}}{\mathrm{24}} \\ $$$${a}_{\mathrm{5}} =−\frac{\mathrm{7}}{\mathrm{15}} \\ $$$${a}_{\mathrm{6}} =−\frac{\mathrm{7}}{\mathrm{10}} \\ $$$${a}_{\mathrm{7}} =−\mathrm{1} \\ $$$${find}\:{a}_{{k}} \\ $$

Commented by Tony Lin last updated on 22/May/20

a_1 =(1/5)C_4 ^5 (−(1/(30))) a_2 =(1/6)C_4 ^6 (−(1/(30))) a_3 =(1/7)C_4 ^7 (−(1/(30))) a_k =(1/(k+4))C_4 ^(k+4) (−(1/(30))) =(((k+4)!)/((k+4)4!k!))(−(1/(30))) =−(((k+1)(k+2)(k+3))/(720))

$${a}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{5}}{C}_{\mathrm{4}} ^{\mathrm{5}} \left(−\frac{\mathrm{1}}{\mathrm{30}}\right) \\ $$$${a}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{6}}{C}_{\mathrm{4}} ^{\mathrm{6}} \left(−\frac{\mathrm{1}}{\mathrm{30}}\right) \\ $$$${a}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{7}}{C}_{\mathrm{4}} ^{\mathrm{7}} \left(−\frac{\mathrm{1}}{\mathrm{30}}\right) \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{{k}+\mathrm{4}}{C}_{\mathrm{4}} ^{{k}+\mathrm{4}} \left(−\frac{\mathrm{1}}{\mathrm{30}}\right) \\ $$$$=\frac{\left({k}+\mathrm{4}\right)!}{\left({k}+\mathrm{4}\right)\mathrm{4}!{k}!}\left(−\frac{\mathrm{1}}{\mathrm{30}}\right) \\ $$$$=−\frac{\left({k}+\mathrm{1}\right)\left({k}+\mathrm{2}\right)\left({k}+\mathrm{3}\right)}{\mathrm{720}} \\ $$

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