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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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