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Question Number 22044 by Tinkutara last updated on 10/Oct/17
$$\mathrm{Let}\:{A}\:=\:\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:.....,\:{n}\right\},\:\mathrm{if}\:{a}_{{i}} \:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{minimum}\:\mathrm{element}\:\mathrm{of}\:\mathrm{the}\:\mathrm{set}\:{A};\:\left(\mathrm{where}\right. \\ $$$${A};\:\mathrm{denotes}\:\mathrm{the}\:\mathrm{subset}\:\mathrm{of}\:{A}\:\mathrm{containing} \\ $$$$\left.\mathrm{exactly}\:\mathrm{three}\:\mathrm{elements}\right)\:\mathrm{and}\:{X}\:\mathrm{denotes} \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:{A}_{{i}} '\mathrm{s},\:\mathrm{then}\:\mathrm{evaluate}\:\underset{{A}_{{i}} \in{X}} {\sum}{a}. \\ $$
Answered by revenge last updated on 17/Oct/17
![Required sum=1∙^(n−1) C_2 +2∙^(n−2) C_2 +3∙^(n−3) C_2 +...+(n−2)∙^2 C_2 It is equal to the coefficient of x^2 in the expansion of (1+x)^(n−1) +2(1+x)^(n−2) +3(1+x)^(n−3) +...+(n−2)(1+x)^2 S=(1+x)^(n−1) +2(1+x)^(n−2) +3(1+x)^(n−3) +...+(n−2)(1+x)^2 (S/(1+x))= (1+x)^(n−2) +2(1+x)^(n−3) +...+(n−3)(1+x)^2 +(n−2)(1+x) S−(S/(1+x))=(1+x)^(n−1) +(1+x)^(n−2) +(1+x)^(n−3) +...+(1+x)^2 −(n−2)(1+x) ((Sx)/(1+x))=(((1+x)^2 [(1+x)^(n−2) −1])/x)−(n−2)(1+x) S=(((1+x)^3 [(1+x)^(n−2) −1])/x^2 )−(((n−2)(1+x)^2 )/x) S=(((1+x)^(n+1) )/x^2 )−(((1+x)^3 )/x^2 )−(n−2)((1+2x+x^2 )/x) x^2 term can be only obtained from (((1+x)^(n+1) )/x^2 ). So we require x^4 in (1+x)^(n+1) , which has the coefficient^(n+1) C_4 .](Q22417.png)
$${Required}\:{sum}=\mathrm{1}\centerdot^{{n}−\mathrm{1}} {C}_{\mathrm{2}} +\mathrm{2}\centerdot^{{n}−\mathrm{2}} {C}_{\mathrm{2}} +\mathrm{3}\centerdot^{{n}−\mathrm{3}} {C}_{\mathrm{2}} +...+\left({n}−\mathrm{2}\right)\centerdot^{\mathrm{2}} {C}_{\mathrm{2}} \\ $$$${It}\:{is}\:{equal}\:{to}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{2}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$$\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} +\mathrm{2}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} +\mathrm{3}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{3}} +...+\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$${S}=\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} +\mathrm{2}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} +\mathrm{3}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{3}} +...+\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$$\frac{{S}}{\mathrm{1}+{x}}=\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} +\mathrm{2}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{3}} +...+\left({n}−\mathrm{3}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} +\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right) \\ $$$${S}−\frac{{S}}{\mathrm{1}+{x}}=\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} +\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} +\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{3}} +...+\left(\mathrm{1}+{x}\right)^{\mathrm{2}} −\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right) \\ $$$$\frac{{Sx}}{\mathrm{1}+{x}}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} \left[\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} −\mathrm{1}\right]}{{x}}−\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right) \\ $$$${S}=\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} \left[\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{2}} −\mathrm{1}\right]}{{x}^{\mathrm{2}} }−\frac{\left({n}−\mathrm{2}\right)\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }{{x}} \\ $$$${S}=\frac{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} }{{x}^{\mathrm{2}} }−\frac{\left(\mathrm{1}+{x}\right)^{\mathrm{3}} }{{x}^{\mathrm{2}} }−\left({n}−\mathrm{2}\right)\frac{\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }{{x}} \\ $$$${x}^{\mathrm{2}} \:{term}\:{can}\:{be}\:{only}\:{obtained}\:{from}\:\frac{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} }{{x}^{\mathrm{2}} }.\:{So}\:{we}\:{require} \\ $$$${x}^{\mathrm{4}} \:{in}\:\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} ,\:{which}\:{has}\:{the}\:{coefficient}\:^{{n}+\mathrm{1}} {C}_{\mathrm{4}} . \\ $$
Commented by Tinkutara last updated on 17/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$