Menu Close

The-coefficient-of-x-4-in-the-expansion-of-1-x-x-2-x-3-11-is-




Question Number 55785 by gunawan last updated on 04/Mar/19
The coefficient of x^4  in the expansion of  (1+x+x^2 +x^3 )^(11)  is
$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:{x}^{\mathrm{4}} \:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)^{\mathrm{11}} \:\mathrm{is} \\ $$
Commented by Tawa1 last updated on 04/Mar/19
coefficient of  x^4   in  (1 + x + x^2  + x^3 )^(11)   is  ...         =  ((11!)/((α_1 )!(α_2 )!(α_3 )!(α_4 )!)) 1^(α_1  ) (x)^α_2   (x^2 )^α_3   (x^3 )^α_4           =  ((11!)/((α_1 )!(α_2 )!(α_3 )!(α_4 )!)) (x)^α_2   (x)^(2α_3 )  (x)^(3α_4 )          =  ((11!)/((α_1 )!(α_2 )!(α_3 )!(α_4 )!)) (x)^(α_2  + 2α_3  + 3α_4 )   where   α_1  , α_2  , α_3  , α_4   are non−negative integer satisfy           α_1  + α_2  + α_3  + α_4   =  11         ......  equation (i)           α_2  + 2α_3  + 3α_4   =  4         ......  equation (ii)  Hence,             α_1  = 9 ,  α_2  = 1 ,  α_3  = 0,  α_4  = 1                               α_1  = 9 ,  α_2  = 0 ,  α_3  = 2,  α_4  = 0                               α_1  = 8 ,  α_2  = 2 ,  α_3  = 1,  α_4  = 0                               α_1  = 7 ,  α_2  = 4 ,  α_3  = 0,  α_4  = 0  coefficient of  x^4   in  (1 + x + x^2  + x^3 )^(11)   is  now           =  ((11!)/(9! 1! 0! 1!))  +  ((11!)/(9! 0! 2! 0!)) + ((11!)/(8! 2! 1! 0!)) + ((11!)/(7! 4! 0! 0!))           =  ((11 × 10 × 9!)/(9! 1! 0! 1!))  +  ((11 × 10 × 9!)/(9! 0! 2! 0!)) + ((11 × 10 × 9 × 8!)/(8! 2! 1! 0!)) + ((11 × 10 × 9 × 8 × 7!)/(7! 4! 0! 0!))           =  110  +  ((110)/(2!)) + ((990)/(2!)) + ((7920)/(4!))           =  110  +  ((110)/2) + ((990)/2) + ((7920)/(24))           =  110 + 55 + 495 + 330           =  990
$$\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{\mathrm{4}} \:\:\mathrm{in}\:\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{11}} \:\:\mathrm{is}\:\:… \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{\mathrm{11}!}{\left(\alpha_{\mathrm{1}} \right)!\left(\alpha_{\mathrm{2}} \right)!\left(\alpha_{\mathrm{3}} \right)!\left(\alpha_{\mathrm{4}} \right)!}\:\mathrm{1}^{\alpha_{\mathrm{1}} \:} \left(\mathrm{x}\right)^{\alpha_{\mathrm{2}} } \:\left(\mathrm{x}^{\mathrm{2}} \right)^{\alpha_{\mathrm{3}} } \:\left(\mathrm{x}^{\mathrm{3}} \right)^{\alpha_{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{\mathrm{11}!}{\left(\alpha_{\mathrm{1}} \right)!\left(\alpha_{\mathrm{2}} \right)!\left(\alpha_{\mathrm{3}} \right)!\left(\alpha_{\mathrm{4}} \right)!}\:\left(\mathrm{x}\right)^{\alpha_{\mathrm{2}} } \:\left(\mathrm{x}\right)^{\mathrm{2}\alpha_{\mathrm{3}} } \:\left(\mathrm{x}\right)^{\mathrm{3}\alpha_{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:=\:\:\frac{\mathrm{11}!}{\left(\alpha_{\mathrm{1}} \right)!\left(\alpha_{\mathrm{2}} \right)!\left(\alpha_{\mathrm{3}} \right)!\left(\alpha_{\mathrm{4}} \right)!}\:\left(\mathrm{x}\right)^{\alpha_{\mathrm{2}} \:+\:\mathrm{2}\alpha_{\mathrm{3}} \:+\:\mathrm{3}\alpha_{\mathrm{4}} } \\ $$$$\mathrm{where}\:\:\:\alpha_{\mathrm{1}} \:,\:\alpha_{\mathrm{2}} \:,\:\alpha_{\mathrm{3}} \:,\:\alpha_{\mathrm{4}} \:\:\mathrm{are}\:\mathrm{non}−\mathrm{negative}\:\mathrm{integer}\:\mathrm{satisfy}\: \\ $$$$\:\:\:\:\:\:\:\:\alpha_{\mathrm{1}} \:+\:\alpha_{\mathrm{2}} \:+\:\alpha_{\mathrm{3}} \:+\:\alpha_{\mathrm{4}} \:\:=\:\:\mathrm{11}\:\:\:\:\:\:\:\:\:……\:\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{2}} \:+\:\mathrm{2}\alpha_{\mathrm{3}} \:+\:\mathrm{3}\alpha_{\mathrm{4}} \:\:=\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:……\:\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{Hence},\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{1}} \:=\:\mathrm{9}\:,\:\:\alpha_{\mathrm{2}} \:=\:\mathrm{1}\:,\:\:\alpha_{\mathrm{3}} \:=\:\mathrm{0},\:\:\alpha_{\mathrm{4}} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{1}} \:=\:\mathrm{9}\:,\:\:\alpha_{\mathrm{2}} \:=\:\mathrm{0}\:,\:\:\alpha_{\mathrm{3}} \:=\:\mathrm{2},\:\:\alpha_{\mathrm{4}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{1}} \:=\:\mathrm{8}\:,\:\:\alpha_{\mathrm{2}} \:=\:\mathrm{2}\:,\:\:\alpha_{\mathrm{3}} \:=\:\mathrm{1},\:\:\alpha_{\mathrm{4}} \:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\alpha_{\mathrm{1}} \:=\:\mathrm{7}\:,\:\:\alpha_{\mathrm{2}} \:=\:\mathrm{4}\:,\:\:\alpha_{\mathrm{3}} \:=\:\mathrm{0},\:\:\alpha_{\mathrm{4}} \:=\:\mathrm{0} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\:\mathrm{x}^{\mathrm{4}} \:\:\mathrm{in}\:\:\left(\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \right)^{\mathrm{11}} \:\:\mathrm{is}\:\:\mathrm{now} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{11}!}{\mathrm{9}!\:\mathrm{1}!\:\mathrm{0}!\:\mathrm{1}!}\:\:+\:\:\frac{\mathrm{11}!}{\mathrm{9}!\:\mathrm{0}!\:\mathrm{2}!\:\mathrm{0}!}\:+\:\frac{\mathrm{11}!}{\mathrm{8}!\:\mathrm{2}!\:\mathrm{1}!\:\mathrm{0}!}\:+\:\frac{\mathrm{11}!}{\mathrm{7}!\:\mathrm{4}!\:\mathrm{0}!\:\mathrm{0}!} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{11}\:×\:\mathrm{10}\:×\:\mathrm{9}!}{\mathrm{9}!\:\mathrm{1}!\:\mathrm{0}!\:\mathrm{1}!}\:\:+\:\:\frac{\mathrm{11}\:×\:\mathrm{10}\:×\:\mathrm{9}!}{\mathrm{9}!\:\mathrm{0}!\:\mathrm{2}!\:\mathrm{0}!}\:+\:\frac{\mathrm{11}\:×\:\mathrm{10}\:×\:\mathrm{9}\:×\:\mathrm{8}!}{\mathrm{8}!\:\mathrm{2}!\:\mathrm{1}!\:\mathrm{0}!}\:+\:\frac{\mathrm{11}\:×\:\mathrm{10}\:×\:\mathrm{9}\:×\:\mathrm{8}\:×\:\mathrm{7}!}{\mathrm{7}!\:\mathrm{4}!\:\mathrm{0}!\:\mathrm{0}!} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\mathrm{110}\:\:+\:\:\frac{\mathrm{110}}{\mathrm{2}!}\:+\:\frac{\mathrm{990}}{\mathrm{2}!}\:+\:\frac{\mathrm{7920}}{\mathrm{4}!} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\mathrm{110}\:\:+\:\:\frac{\mathrm{110}}{\mathrm{2}}\:+\:\frac{\mathrm{990}}{\mathrm{2}}\:+\:\frac{\mathrm{7920}}{\mathrm{24}} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\mathrm{110}\:+\:\mathrm{55}\:+\:\mathrm{495}\:+\:\mathrm{330} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\mathrm{990} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19
{(1+x)+x^2 (1+x)}^(11)   {(1+x)(1+x^2 )}^(11)   (1+x)^(11) (1+x^2 )^(11)   (1+11c_1 x+11c_2 x^2 +11c_3 x^3 +11c_4 x^4 +...+11c_(11) x^(11) )×(1+11c_1 x^2 +11c_2 x^4 +11c_3 x^6 +...+11c_(11) x^(22) )  terms containing x^4  are  1×11c_2 x^4 +11c_2 x^2 ×11c_1 x^2 +11c_4 x^4 ×1  =x^4 (11c_2 +11c_2 ×11c_1 +11c_4 )  =x^4 (((11!)/(2!9!))+((11!)/(2!9!))×((11!)/(1!10!))+((11!)/(4!7!)))  =x^4 (55+55×11+((11×10×9×8)/(4×3×2)))  =x^4 (55+605+330)  =x^4 (990)  pls check...
$$\left\{\left(\mathrm{1}+{x}\right)+{x}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right\}^{\mathrm{11}} \\ $$$$\left\{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right\}^{\mathrm{11}} \\ $$$$\left(\mathrm{1}+{x}\right)^{\mathrm{11}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{11}} \\ $$$$\left(\mathrm{1}+\mathrm{11}{c}_{\mathrm{1}} {x}+\mathrm{11}{c}_{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{11}{c}_{\mathrm{3}} {x}^{\mathrm{3}} +\mathrm{11}{c}_{\mathrm{4}} {x}^{\mathrm{4}} +…+\mathrm{11}{c}_{\mathrm{11}} {x}^{\mathrm{11}} \right)×\left(\mathrm{1}+\mathrm{11}{c}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{11}{c}_{\mathrm{2}} {x}^{\mathrm{4}} +\mathrm{11}{c}_{\mathrm{3}} {x}^{\mathrm{6}} +…+\mathrm{11}{c}_{\mathrm{11}} {x}^{\mathrm{22}} \right) \\ $$$${terms}\:{containing}\:{x}^{\mathrm{4}} \:{are} \\ $$$$\mathrm{1}×\mathrm{11}{c}_{\mathrm{2}} {x}^{\mathrm{4}} +\mathrm{11}{c}_{\mathrm{2}} {x}^{\mathrm{2}} ×\mathrm{11}{c}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{11}{c}_{\mathrm{4}} {x}^{\mathrm{4}} ×\mathrm{1} \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{11}{c}_{\mathrm{2}} +\mathrm{11}{c}_{\mathrm{2}} ×\mathrm{11}{c}_{\mathrm{1}} +\mathrm{11}{c}_{\mathrm{4}} \right) \\ $$$$={x}^{\mathrm{4}} \left(\frac{\mathrm{11}!}{\mathrm{2}!\mathrm{9}!}+\frac{\mathrm{11}!}{\mathrm{2}!\mathrm{9}!}×\frac{\mathrm{11}!}{\mathrm{1}!\mathrm{10}!}+\frac{\mathrm{11}!}{\mathrm{4}!\mathrm{7}!}\right) \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{55}+\mathrm{55}×\mathrm{11}+\frac{\mathrm{11}×\mathrm{10}×\mathrm{9}×\mathrm{8}}{\mathrm{4}×\mathrm{3}×\mathrm{2}}\right) \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{55}+\mathrm{605}+\mathrm{330}\right) \\ $$$$={x}^{\mathrm{4}} \left(\mathrm{990}\right) \\ $$$${pls}\:{check}… \\ $$
Answered by mr W last updated on 06/Mar/19
1+x+x^2 +x^3 =((1−x^4 )/(1−x))  (1+x+x^2 +x^3 )^(11)   =(1−x^4 )^(11) (1−x)^(−11)   =Σ_(k=0) ^(11) C_k ^(11) (−x^4 )^k Σ_(k=0) ^∞ C_k ^(10+k) x^k   coef. of x^4 :  C_1 ^(11) (−1)^1 +C_4 ^(14)   =−11+1001  =990
$$\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} =\frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}−{x}} \\ $$$$\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} \right)^{\mathrm{11}} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{4}} \right)^{\mathrm{11}} \left(\mathrm{1}−{x}\right)^{−\mathrm{11}} \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\mathrm{11}} {\sum}}{C}_{{k}} ^{\mathrm{11}} \left(−{x}^{\mathrm{4}} \right)^{{k}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{k}} ^{\mathrm{10}+{k}} {x}^{{k}} \\ $$$${coef}.\:{of}\:{x}^{\mathrm{4}} : \\ $$$${C}_{\mathrm{1}} ^{\mathrm{11}} \left(−\mathrm{1}\right)^{\mathrm{1}} +{C}_{\mathrm{4}} ^{\mathrm{14}} \\ $$$$=−\mathrm{11}+\mathrm{1001} \\ $$$$=\mathrm{990} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 05/Mar/19
sum of gp series...S=((a(1−r^n ))/(1−r))=((1−x^4 )/(1−x))
$${sum}\:{of}\:{gp}\:{series}…{S}=\frac{{a}\left(\mathrm{1}−{r}^{{n}} \right)}{\mathrm{1}−{r}}=\frac{\mathrm{1}−{x}^{\mathrm{4}} }{\mathrm{1}−{x}} \\ $$
Commented by Tawa1 last updated on 05/Mar/19
Sir, please can you explain this method very well sir,   i like it sir.   How    1 + x + x^2  + x^3  = ((1 − x^4 )/(1 − x))   and the summation part ..  etc ...
$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{can}\:\mathrm{you}\:\mathrm{explain}\:\mathrm{this}\:\mathrm{method}\:\mathrm{very}\:\mathrm{well}\:\mathrm{sir},\: \\ $$$$\mathrm{i}\:\mathrm{like}\:\mathrm{it}\:\mathrm{sir}.\:\:\:\mathrm{How}\:\:\:\:\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{4}} }{\mathrm{1}\:−\:\mathrm{x}}\:\:\:\mathrm{and}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{part}\:.. \\ $$$$\mathrm{etc}\:… \\ $$
Commented by Tawa1 last updated on 05/Mar/19
God bless you sir. i appreciate
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{appreciate} \\ $$
Commented by Tawa1 last updated on 05/Mar/19
Why is the summation zero to infinity for power −11
$$\mathrm{Why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{summation}\:\mathrm{zero}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{for}\:\mathrm{power}\:−\mathrm{11} \\ $$
Commented by mr W last updated on 06/Mar/19
(1/(1−x))=1+x+x^2 +x^3 +... →infinite  (1/((1−x)^p ))=(1+x+x^2 +x^3 +...)^p →infinite  (1/((1−x)^p ))=Σ_(k=0) ^∞ a_k x^k
$$\frac{\mathrm{1}}{\mathrm{1}−{x}}=\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\:\rightarrow{infinite} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{{p}} }=\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +…\right)^{{p}} \rightarrow{infinite} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{{p}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{k}} {x}^{{k}} \\ $$
Commented by Tawa1 last updated on 06/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 06/Mar/19
it′s useful:  (1/((1−x)^p ))=Σ_(k=0) ^∞ C_k ^( p−1+k) x^k
$${it}'{s}\:{useful}: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{{p}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{k}} ^{\:{p}−\mathrm{1}+{k}} {x}^{{k}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *