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Question-57044

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Question Number 57044 by mr W last updated on 29/Mar/19
Commented by mr W last updated on 29/Mar/19
reposted Question from Tinkutara sir
$${reposted} \\ $$$${Question}\:{from}\:{Tinkutara}\:{sir} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Mar/19
number of A type object=2n numbser of B type object=2n number of C type object=2n so total object=6n to be distributed between two person. (1+x+x^2 +..+x^(2n) )(1+x+x^2 +...+x^(2n) )(1+x+x^2 +..x^(2n) ) ((1−x^(2n+1) )/(1−x))×((1−x^(2n+1) )/(1−x))×((1−x^(2n+1) )/(1−x)) =(((1−x^(2n+1) )^3 )/((1−x)^3 )) =(1−x)^(−3) (1−x^(2n+1) )^3 required answer is the coefficient of x^(3n) in (1−x)^(−3) (1−x^(2n+1) )^3 =(1+3x+6x^2 +..+(((r+1)(r+2))/(1×2))x^r +..)(1−3x^(2n+1) +3x^(4n+2) −x^(6n+3) ) =(((3n+1)(3n+2))/(1×2))x^(3n) −3x^(2n+1) ×(((n−1+1)(n−1+2))/(1×2))x^(n−1) =x^(3n) {((9n^2 +9n+2)/2)−(3/2)×(n^2 +n)} =(x^(3n) /2)(9n^2 +9n+2−3n^2 −3n) =(x^(3n) /2)(6n^2 +6n+2) so answer is 3n^2 +3n+1
$${number}\:{of}\:{A}\:{type}\:{object}=\mathrm{2}{n} \\ $$$${numbser}\:{of}\:{B}\:{type}\:{object}=\mathrm{2}{n} \\ $$$${number}\:{of}\:{C}\:{ty}\boldsymbol{{p}}{e}\:{object}=\mathrm{2}{n} \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{total}}\:\boldsymbol{{object}}=\mathrm{6}\boldsymbol{{n}} \\ $$$$\boldsymbol{{to}}\:\boldsymbol{{be}}\:\boldsymbol{{distributed}}\:\boldsymbol{{between}}\:\boldsymbol{{two}}\:\boldsymbol{{person}}. \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +..+{x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +…+{x}^{\mathrm{2}{n}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +..{x}^{\mathrm{2}{n}} \right) \\ $$$$\frac{\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}}×\frac{\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}}×\frac{\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{1}−{x}} \\ $$$$=\frac{\left(\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} \right)^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$=\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} \left(\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} \right)^{\mathrm{3}} \\ $$$$\boldsymbol{{required}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\boldsymbol{{the}}\:\boldsymbol{{coefficient}}\:\boldsymbol{{of}}\: \\ $$$$\boldsymbol{{x}}^{\mathrm{3}\boldsymbol{{n}}} \:\boldsymbol{{in}}\:\left(\mathrm{1}−\boldsymbol{{x}}\right)^{−\mathrm{3}} \left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}\boldsymbol{{n}}+\mathrm{1}} \right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{1}+\mathrm{3}\boldsymbol{{x}}+\mathrm{6}\boldsymbol{{x}}^{\mathrm{2}} +..+\frac{\left(\boldsymbol{{r}}+\mathrm{1}\right)\left(\boldsymbol{{r}}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{{r}} +..\right)\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}{x}^{\mathrm{4}{n}+\mathrm{2}} −{x}^{\mathrm{6}{n}+\mathrm{3}} \right) \\ $$$$=\frac{\left(\mathrm{3}{n}+\mathrm{1}\right)\left(\mathrm{3}{n}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{\mathrm{3}{n}} −\mathrm{3}{x}^{\mathrm{2}{n}+\mathrm{1}} ×\frac{\left({n}−\mathrm{1}+\mathrm{1}\right)\left({n}−\mathrm{1}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{{n}−\mathrm{1}} \\ $$$$={x}^{\mathrm{3}{n}} \left\{\frac{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{9}{n}+\mathrm{2}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}×\left({n}^{\mathrm{2}} +{n}\right)\right\} \\ $$$$=\frac{{x}^{\mathrm{3}{n}} }{\mathrm{2}}\left(\mathrm{9}{n}^{\mathrm{2}} +\mathrm{9}{n}+\mathrm{2}−\mathrm{3}{n}^{\mathrm{2}} −\mathrm{3}{n}\right) \\ $$$$=\frac{{x}^{\mathrm{3}{n}} }{\mathrm{2}}\left(\mathrm{6}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{2}\right) \\ $$$${so}\:{answer}\:{is}\:\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 30/Mar/19
thanks sir! nice proof! (((1−x^(2n+1) )^3 )/((1−x)^3 )) =(1−3x^(2n+1) +3x^(4n+2) −x^(6n+3) )Σ_(k=0) ^∞ C_k ^(2+k) x^k x^(3n) : C_(3n) ^(2+3n) −3C_(n−1) ^(n+1) =C_2 ^(3n+2) −3C_2 ^(n+1) =(((3n+2)(3n+1)−3(n+1)n)/2) =((6n^2 +6n+2)/2) =3n^2 +3n+1
$${thanks}\:{sir}!\:{nice}\:{proof}! \\ $$$$ \\ $$$$\frac{\left(\mathrm{1}−{x}^{\mathrm{2}{n}+\mathrm{1}} \right)^{\mathrm{3}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{3}} } \\ $$$$=\left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}{n}+\mathrm{1}} +\mathrm{3}{x}^{\mathrm{4}{n}+\mathrm{2}} −{x}^{\mathrm{6}{n}+\mathrm{3}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{k}} ^{\mathrm{2}+{k}} {x}^{{k}} \\ $$$${x}^{\mathrm{3}{n}} :\:{C}_{\mathrm{3}{n}} ^{\mathrm{2}+\mathrm{3}{n}} −\mathrm{3}{C}_{{n}−\mathrm{1}} ^{{n}+\mathrm{1}} ={C}_{\mathrm{2}} ^{\mathrm{3}{n}+\mathrm{2}} −\mathrm{3}{C}_{\mathrm{2}} ^{{n}+\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{3}{n}+\mathrm{2}\right)\left(\mathrm{3}{n}+\mathrm{1}\right)−\mathrm{3}\left({n}+\mathrm{1}\right){n}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{6}{n}^{\mathrm{2}} +\mathrm{6}{n}+\mathrm{2}}{\mathrm{2}} \\ $$$$=\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 30/Mar/19
this platform bind us by invisble bond and boost us to quench our thirst of quest to find the truth... the solution...
$${this}\:{platform}\:{bind}\:{us}\:{by}\:{invisble}\:{bond}\:{and}\:{boost} \\ $$$${us}\:{to}\:{quench}\:{our}\:{thirst}\:{of}\:{quest}\:{to}\:{find}\:{the}\:{truth}… \\ $$$${the}\:{solution}… \\ $$

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