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Question-104497




Question Number 104497 by PengagumRahasiamu last updated on 21/Jul/20
Commented by mr W last updated on 22/Jul/20
Q2:  a≤2  ?
$${Q}\mathrm{2}: \\ $$$${a}\leqslant\mathrm{2}\:\:? \\ $$
Commented by PengagumRahasiamu last updated on 22/Jul/20
Non-negative integer Sir
Answered by mr W last updated on 22/Jul/20
suppose a,b,...∈N  Q1:  (x+x^2 +x^3 +x^4 +x^5 +x^6 )^4   =x^4 (1+x+x^2 +x^3 +x^4 +x^5 )^4   =((x^4 (1−x^6 )^4 )/((1−x)^4 ))  =x^4 (1−4x^6 +6x^(12) −4x^(18) +x^(24) )Σ_(k=0) ^∞ C_3 ^(k+3) x^k   coefficient of x^(20)  term:  C_3 ^(19) −4×C_3 ^(13) +6×C_3 ^7 =35  ⇒35 tuples    Q2:  0 tuples if a≤2  since 2+3+4+5+6+7=27<30  2520 tuples if a≥1
$${suppose}\:{a},{b},…\in\mathbb{N} \\ $$$${Q}\mathrm{1}: \\ $$$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +{x}^{\mathrm{6}} \right)^{\mathrm{4}} \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)^{\mathrm{4}} \\ $$$$=\frac{{x}^{\mathrm{4}} \left(\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{4}} }{\left(\mathrm{1}−{x}\right)^{\mathrm{4}} } \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{4}{x}^{\mathrm{6}} +\mathrm{6}{x}^{\mathrm{12}} −\mathrm{4}{x}^{\mathrm{18}} +{x}^{\mathrm{24}} \right)\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{\mathrm{3}} ^{{k}+\mathrm{3}} {x}^{{k}} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{20}} \:{term}: \\ $$$${C}_{\mathrm{3}} ^{\mathrm{19}} −\mathrm{4}×{C}_{\mathrm{3}} ^{\mathrm{13}} +\mathrm{6}×{C}_{\mathrm{3}} ^{\mathrm{7}} =\mathrm{35} \\ $$$$\Rightarrow\mathrm{35}\:{tuples} \\ $$$$ \\ $$$${Q}\mathrm{2}: \\ $$$$\mathrm{0}\:{tuples}\:{if}\:{a}\leqslant\mathrm{2} \\ $$$${since}\:\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}=\mathrm{27}<\mathrm{30} \\ $$$$\mathrm{2520}\:{tuples}\:{if}\:{a}\geqslant\mathrm{1} \\ $$

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