Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 8217 by tawakalitu last updated on 02/Oct/16

what is the coefficient of x^3 in the expansion of (1 + x + x^2 + x^3 + x^4 + x^5 )^6

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{x}^{\mathrm{5}} \right)^{\mathrm{6}} \\ $$

Commented by 123456 last updated on 04/Oct/16

1∙2∙3∙4∙5∙6 1+1+1=1+2=3 3:6 1+2:6∙5=30 1+1+1:((6∙5∙4)/(3!))=((6∙5∙4)/(3∙2))=5∙4=20 6+30+20=56

$$\mathrm{1}\centerdot\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\mathrm{3}:\mathrm{6} \\ $$$$\mathrm{1}+\mathrm{2}:\mathrm{6}\centerdot\mathrm{5}=\mathrm{30} \\ $$$$\mathrm{1}+\mathrm{1}+\mathrm{1}:\frac{\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}}{\mathrm{3}!}=\frac{\mathrm{6}\centerdot\mathrm{5}\centerdot\mathrm{4}}{\mathrm{3}\centerdot\mathrm{2}}=\mathrm{5}\centerdot\mathrm{4}=\mathrm{20} \\ $$$$\mathrm{6}+\mathrm{30}+\mathrm{20}=\mathrm{56} \\ $$

Commented by sandy_suhendra last updated on 04/Oct/16

this is so simple, but can you more explain it?

$$\mathrm{this}\:\mathrm{is}\:\mathrm{so}\:\mathrm{simple},\:\mathrm{but}\:\mathrm{can}\:\mathrm{you}\:\mathrm{more}\:\mathrm{explain}\:\mathrm{it}? \\ $$

Commented by 123456 last updated on 04/Oct/16

we have (1+x+x^2 +x^3 +x^4 +x^5 )^6 = (1+∙∙∙+x^5 )...(1+∙∙∙+x^5 ) recal x^3 =x^2 x=xxx we have to pick one factor from each (), if we pick x^3 , all other will be 1 we have 6 () so 6 ways if we pick x^2 , we need to pick a x and all other will be 1, so 6() × 5() = 30 ways if we pick x, we pick x^2 (above case) or x and x, so 6 () × 5() × 4() =“120 ways” however x×x×x=x×x×x (3!=6 ways of perm x) so ((120)/6)=20 ways since all coef are 1 the coef is the sum of number of way 6+30+20=56

$$\mathrm{we}\:\mathrm{have} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)^{\mathrm{6}} = \\ $$$$\left(\mathrm{1}+\centerdot\centerdot\centerdot+{x}^{\mathrm{5}} \right)...\left(\mathrm{1}+\centerdot\centerdot\centerdot+{x}^{\mathrm{5}} \right) \\ $$$$\mathrm{recal}\:{x}^{\mathrm{3}} ={x}^{\mathrm{2}} {x}={xxx} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{pick}\:\mathrm{one}\:\mathrm{factor}\:\mathrm{from}\:\mathrm{each} \\ $$$$\left(\right),\:\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{3}} ,\:\mathrm{all}\:\mathrm{other}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{6}\:\left(\right)\:\mathrm{so}\:\mathrm{6}\:\mathrm{ways} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{2}} ,\:\mathrm{we}\:\mathrm{need}\:\mathrm{to}\:\mathrm{pick}\:\mathrm{a}\:{x}\:\mathrm{and} \\ $$$$\mathrm{all}\:\mathrm{other}\:\mathrm{will}\:\mathrm{be}\:\mathrm{1},\:\mathrm{so} \\ $$$$\mathrm{6}\left(\right)\:×\:\mathrm{5}\left(\right)\:=\:\mathrm{30}\:\mathrm{ways} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{pick}\:{x},\:\mathrm{we}\:\mathrm{pick}\:{x}^{\mathrm{2}} \:\left(\mathrm{above}\:\mathrm{case}\right) \\ $$$$\mathrm{or}\:{x}\:\mathrm{and}\:{x},\:\mathrm{so} \\ $$$$\mathrm{6}\:\left(\right)\:×\:\mathrm{5}\left(\right)\:×\:\mathrm{4}\left(\right)\:=``\mathrm{120}\:\mathrm{ways}'' \\ $$$$\mathrm{however} \\ $$$${x}×{x}×{x}={x}×{x}×{x}\:\left(\mathrm{3}!=\mathrm{6}\:\mathrm{ways}\:\mathrm{of}\:\mathrm{perm}\:{x}\right) \\ $$$$\mathrm{so} \\ $$$$\frac{\mathrm{120}}{\mathrm{6}}=\mathrm{20}\:\mathrm{ways} \\ $$$$\mathrm{since}\:\mathrm{all}\:\mathrm{coef}\:\mathrm{are}\:\mathrm{1}\:\mathrm{the}\:\mathrm{coef}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum} \\ $$$$\mathrm{of}\:\mathrm{number}\:\mathrm{of}\:\mathrm{way} \\ $$$$\mathrm{6}+\mathrm{30}+\mathrm{20}=\mathrm{56} \\ $$

Commented by sandy_suhendra last updated on 04/Oct/16

great! thank′s for your answer

$$\mathrm{great}!\:\mathrm{thank}'\mathrm{s}\:\mathrm{for}\:\mathrm{your}\:\mathrm{answer} \\ $$

Commented by tawakalitu last updated on 04/Oct/16

Thanks so much. i really appreciate.

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}.\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Answered by sandy_suhendra last updated on 03/Oct/16

[(1+x+x^2 )+x^3 (1+x+x^2 )]^6 =[(x^3 +1)(x^2 +x+1)]^6 =(x^3 +1)^6 (x^2 +x+1)^6 (x^3 +1)^6 =(x^3 )^6 +6(x^3 )^5 +...+6x^3 +1 [x^2 +(x+1)]^6 =1(x^2 )^6 +6(x^2 )^5 (x+1)+...+6x^2 (x+1)^5 +1(x+1)^6 6x^2 (x+1)^5 =6x^2 (x^5 +5x^4 +...+5x+1)⇒30x^3 (x+1)^6 =x^6 +6x^5 +...+20x^3 +...+1 the term of x^3 comes from ⇒ 6x^3 +30x^3 +20x^3 =56x^3 so the coefficient of x^3 is 56

$$\left[\left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)+\mathrm{x}^{\mathrm{3}} \left(\mathrm{1}+\mathrm{x}+\mathrm{x}^{\mathrm{2}} \right)\right]^{\mathrm{6}} \\ $$$$=\left[\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)\right]^{\mathrm{6}} \\ $$$$=\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{6}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\left(\mathrm{x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{6}} =\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{6}} +\mathrm{6}\left(\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{5}} +...+\mathrm{6x}^{\mathrm{3}} +\mathrm{1} \\ $$$$\left[\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{1}\right)\right]^{\mathrm{6}} \\ $$$$=\mathrm{1}\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{6}} +\mathrm{6}\left(\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{5}} \left(\mathrm{x}+\mathrm{1}\right)+...+\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} +\mathrm{1}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} \\ $$$$\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} =\mathrm{6x}^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{5}} +\mathrm{5x}^{\mathrm{4}} +...+\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{1}\right)\Rightarrow\mathrm{30x}^{\mathrm{3}} \\ $$$$\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{6}} =\mathrm{x}^{\mathrm{6}} +\mathrm{6x}^{\mathrm{5}} +...+\mathrm{20x}^{\mathrm{3}} +...+\mathrm{1} \\ $$$$\mathrm{the}\:\mathrm{term}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{comes}\:\mathrm{from}\:\Rightarrow\:\mathrm{6x}^{\mathrm{3}} +\mathrm{30x}^{\mathrm{3}} +\mathrm{20x}^{\mathrm{3}} =\mathrm{56x}^{\mathrm{3}} \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} \:\mathrm{is}\:\mathrm{56} \\ $$

Commented by tawakalitu last updated on 03/Oct/16

Thanks so much.

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com