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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
$${it}'{s}\:{useful}: \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{{p}} }=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{C}_{{k}} ^{\:{p}−\mathrm{1}+{k}} {x}^{{k}} \\ $$

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