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Question Number 9573 by lepan last updated on 17/Dec/16
$${If}\:{n}\:{is}\:{positive}\:{integer}\:{prove}\:{that}\: \\ $$$${the}\:{cofficient}\:{of}\:{x}^{\mathrm{2}\:} {and}\:{x}^{\mathrm{3}} \:{in}\:{the}\: \\ $$$${expansion}\:{of}\:\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)^{{n}} \:{are}\:\mathrm{2}^{{n}−\mathrm{1}} .{n}^{\mathrm{2}} \\ $$$${and}\:\mathrm{2}^{{n}−\mathrm{1}} {n}\left({n}−\mathrm{1}\right)\frac{\mathrm{1}}{\mathrm{3}}. \\ $$
Commented by sou1618 last updated on 17/Dec/16
![a,b,c=0,1,2,3.... m(,n)=1,2,3.... kx^m =Σ_((a,b,c)) {(x^2 )^a ×(2x)^b ×2^c × _n C_a × _(n−a) C_b ×_(n−a−b) C_c } k=cofficient of x^m m=2a+b+0c a+b+c=n _(n−a−b) C_c = _c C_c =1 // // // [kx^2 ] m=2 ⇒ (a,b,c)=^((1)) (1,0,n−1),^((2)) (0,2,n−2) (1)k_1 x^2 =(x^2 )^1 ×(2x)^0 ×(2)^(n−1) ×_n C_1 ×_(n−1) C_0 ⇒k_1 =2^(n−1) ×n (2)k_2 x^2 =(x^2 )^0 ×(2x)^2 ×2^(n−2) ×_n C_0 ×_n C_2 ⇒k_2 =2^n ×((n(n−1))/2)=2^(n−1) (n^2 −n) ⇒k=k_1 +k_2 =2^(n−1) (n^2 −n+n) =2^(n−1) n^2 // // // [kx^3 ] m=3 ⇒ (a,b,c)=^((1)) (1,1,n−2),^((2)) (0,3,n−3) (1)k_1 x^3 =(x^2 )^1 ×(2x)^1 ×2^(n−2) ×_n C_1 ×_(n−1) C_1 ⇒k_1 =2×2^(n−2) ×n×(n−1)=2^(n−1) (n^2 −n) (2)k_2 x^3 =(x^2 )^0 ×(2x)^3 ×2^(n−3) ×_n C_0 ×_n C_3 ⇒k_2 =2^3 ×2^(n−3) ×((n(n−1)(n−2))/6)=2^(n−1) (((n^2 −n)(n−2))/3) ⇒k=k_1 +k_2 =2^(n−1) (n^2 −n)(1+((n−2)/3)) =2^(n−1) (((n−1)n(n+1))/3)](Q9574.png)
$${a},{b},{c}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}....\:{m}\left(,{n}\right)=\mathrm{1},\mathrm{2},\mathrm{3}.... \\ $$$$ \\ $$$${kx}^{{m}} =\underset{\left({a},{b},{c}\right)} {\sum}\left\{\left({x}^{\mathrm{2}} \right)^{{a}} ×\left(\mathrm{2}{x}\right)^{{b}} ×\mathrm{2}^{{c}} ×\:_{{n}} {C}_{{a}} ×\:_{{n}−{a}} {C}_{{b}} ×_{{n}−{a}−{b}} {C}_{{c}} \right\} \\ $$$$\:\:{k}={cofficient}\:{of}\:{x}^{{m}} \\ $$$$\:\:{m}=\mathrm{2}{a}+{b}+\mathrm{0}{c} \\ $$$$\:\:{a}+{b}+{c}={n} \\ $$$$ \\ $$$$\:_{{n}−{a}−{b}} {C}_{{c}} =\:_{{c}} {C}_{{c}} =\mathrm{1} \\ $$$$ \\ $$$$//\://\://\: \\ $$$$\left[{kx}^{\mathrm{2}} \right] \\ $$$$\:\:{m}=\mathrm{2}\:\Rightarrow\:\left({a},{b},{c}\right)=\:^{\left(\mathrm{1}\right)} \left(\mathrm{1},\mathrm{0},{n}−\mathrm{1}\right),\:^{\left(\mathrm{2}\right)} \left(\mathrm{0},\mathrm{2},{n}−\mathrm{2}\right) \\ $$$$\:\:\left(\mathrm{1}\right){k}_{\mathrm{1}} {x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{1}} ×\left(\mathrm{2}{x}\right)^{\mathrm{0}} ×\left(\mathrm{2}\right)^{{n}−\mathrm{1}} ×_{{n}} {C}_{\mathrm{1}} ×_{{n}−\mathrm{1}} {C}_{\mathrm{0}} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{1}} =\mathrm{2}^{{n}−\mathrm{1}} ×{n} \\ $$$$\:\:\left(\mathrm{2}\right){k}_{\mathrm{2}} {x}^{\mathrm{2}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{0}} ×\left(\mathrm{2}{x}\right)^{\mathrm{2}} ×\mathrm{2}^{{n}−\mathrm{2}} ×_{{n}} {C}_{\mathrm{0}} ×_{{n}} {C}_{\mathrm{2}} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{2}} =\mathrm{2}^{{n}} ×\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{2}^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{n}\right) \\ $$$$\:\Rightarrow{k}={k}_{\mathrm{1}} +{k}_{\mathrm{2}} =\mathrm{2}^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{n}+{n}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}^{{n}−\mathrm{1}} {n}^{\mathrm{2}} \\ $$$$ \\ $$$$//\://\:// \\ $$$$\left[{kx}^{\mathrm{3}} \right] \\ $$$$\:\:{m}=\mathrm{3}\:\Rightarrow\:\left({a},{b},{c}\right)=\:^{\left(\mathrm{1}\right)} \left(\mathrm{1},\mathrm{1},{n}−\mathrm{2}\right),\:^{\left(\mathrm{2}\right)} \left(\mathrm{0},\mathrm{3},{n}−\mathrm{3}\right) \\ $$$$\:\:\left(\mathrm{1}\right){k}_{\mathrm{1}} {x}^{\mathrm{3}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{1}} ×\left(\mathrm{2}{x}\right)^{\mathrm{1}} ×\mathrm{2}^{{n}−\mathrm{2}} ×_{{n}} {C}_{\mathrm{1}} ×_{{n}−\mathrm{1}} {C}_{\mathrm{1}} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{1}} =\mathrm{2}×\mathrm{2}^{{n}−\mathrm{2}} ×{n}×\left({n}−\mathrm{1}\right)=\mathrm{2}^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{n}\right) \\ $$$$\:\:\left(\mathrm{2}\right){k}_{\mathrm{2}} {x}^{\mathrm{3}} =\left({x}^{\mathrm{2}} \right)^{\mathrm{0}} ×\left(\mathrm{2}{x}\right)^{\mathrm{3}} ×\mathrm{2}^{{n}−\mathrm{3}} ×_{{n}} {C}_{\mathrm{0}} ×_{{n}} {C}_{\mathrm{3}} \\ $$$$\:\:\:\:\Rightarrow{k}_{\mathrm{2}} =\mathrm{2}^{\mathrm{3}} ×\mathrm{2}^{{n}−\mathrm{3}} ×\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{6}}=\mathrm{2}^{{n}−\mathrm{1}} \frac{\left({n}^{\mathrm{2}} −{n}\right)\left({n}−\mathrm{2}\right)}{\mathrm{3}}\: \\ $$$$\:\Rightarrow{k}={k}_{\mathrm{1}} +{k}_{\mathrm{2}} =\mathrm{2}^{{n}−\mathrm{1}} \left({n}^{\mathrm{2}} −{n}\right)\left(\mathrm{1}+\frac{{n}−\mathrm{2}}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}^{{n}−\mathrm{1}} \frac{\left({n}−\mathrm{1}\right){n}\left({n}+\mathrm{1}\right)}{\mathrm{3}} \\ $$$$ \\ $$