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Question Number 5003 by 314159 last updated on 30/Mar/16
Find the coefficient of x^(17)  in the expansion of  (x+1)(x+2)(x+3)...(x+19).
$${Find}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{17}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+\mathrm{19}\right). \\ $$
Commented by 123456 last updated on 30/Mar/16
1+1+...+1 (17)  1+1+...+1 (17)+0  1+1+...+1 (17)+0+0  ((19!)/(17!(19−17)!))=((19∙18)/6)=19∙3=57  (57+1)+...+(57+19)=1083+190=1273  (57+1+2)+(57+1+3)...+(57+18+19)  ((19!)/(2!(19−2)!))=171
$$\mathrm{1}+\mathrm{1}+…+\mathrm{1}\:\left(\mathrm{17}\right) \\ $$$$\mathrm{1}+\mathrm{1}+…+\mathrm{1}\:\left(\mathrm{17}\right)+\mathrm{0} \\ $$$$\mathrm{1}+\mathrm{1}+…+\mathrm{1}\:\left(\mathrm{17}\right)+\mathrm{0}+\mathrm{0} \\ $$$$\frac{\mathrm{19}!}{\mathrm{17}!\left(\mathrm{19}−\mathrm{17}\right)!}=\frac{\mathrm{19}\centerdot\mathrm{18}}{\mathrm{6}}=\mathrm{19}\centerdot\mathrm{3}=\mathrm{57} \\ $$$$\left(\mathrm{57}+\mathrm{1}\right)+…+\left(\mathrm{57}+\mathrm{19}\right)=\mathrm{1083}+\mathrm{190}=\mathrm{1273} \\ $$$$\left(\mathrm{57}+\mathrm{1}+\mathrm{2}\right)+\left(\mathrm{57}+\mathrm{1}+\mathrm{3}\right)…+\left(\mathrm{57}+\mathrm{18}+\mathrm{19}\right) \\ $$$$\frac{\mathrm{19}!}{\mathrm{2}!\left(\mathrm{19}−\mathrm{2}\right)!}=\mathrm{171} \\ $$
Commented by prakash jain last updated on 31/Mar/16
  Coefficient of x^(17)  can be generated by multiplication  any 2 constant terms and remaining x terms.  =1∙2+1∙3+...+1∙19  +2.3+2.4+..+2.19  +  ...  +18∙19  =Σ_(i=1) ^(18)  i(Σ_(j=i+1) ^(19) j)  Arithmetic Progression formulas to sum.
$$ \\ $$$$\mathrm{Coefficient}\:\mathrm{of}\:{x}^{\mathrm{17}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{generated}\:\mathrm{by}\:\mathrm{multiplication} \\ $$$$\mathrm{any}\:\mathrm{2}\:\mathrm{constant}\:\mathrm{terms}\:\mathrm{and}\:\mathrm{remaining}\:{x}\:\mathrm{terms}. \\ $$$$=\mathrm{1}\centerdot\mathrm{2}+\mathrm{1}\centerdot\mathrm{3}+…+\mathrm{1}\centerdot\mathrm{19} \\ $$$$+\mathrm{2}.\mathrm{3}+\mathrm{2}.\mathrm{4}+..+\mathrm{2}.\mathrm{19} \\ $$$$+ \\ $$$$… \\ $$$$+\mathrm{18}\centerdot\mathrm{19} \\ $$$$=\underset{{i}=\mathrm{1}} {\overset{\mathrm{18}} {\sum}}\:{i}\left(\underset{{j}={i}+\mathrm{1}} {\overset{\mathrm{19}} {\sum}}{j}\right) \\ $$$$\mathrm{Arithmetic}\:\mathrm{Progression}\:\mathrm{formulas}\:\mathrm{to}\:\mathrm{sum}. \\ $$

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