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Question Number 82658 by jagoll last updated on 23/Feb/20

coefficient x^6 from expressi (2x+1)^(6 ) × (x^2 +x+(1/4))^4 ?

$${coefficient}\:{x}^{\mathrm{6}} \:{from}\:{expressi}\: \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{6}\:} ×\:\left({x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \:? \\ $$

Commented by mr W last updated on 23/Feb/20

=2^6 (x+(1/2))^6 (x+(1/2))^8 =2^6 (x+(1/2))^(14) =2^6 Σ(1/2^(14−k) )C_k ^(14) x^k k=6: 2^6 ×(1/2^(14−6) )×C_6 ^(14) =(C_6 ^(14) /2^2 )=((3003)/4)

$$=\mathrm{2}^{\mathrm{6}} \left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{6}} \left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{8}} \\ $$$$=\mathrm{2}^{\mathrm{6}} \left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{14}} \\ $$$$=\mathrm{2}^{\mathrm{6}} \Sigma\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{14}−{k}} }{C}_{{k}} ^{\mathrm{14}} {x}^{{k}} \\ $$$${k}=\mathrm{6}: \\ $$$$\mathrm{2}^{\mathrm{6}} ×\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{14}−\mathrm{6}} }×{C}_{\mathrm{6}} ^{\mathrm{14}} =\frac{{C}_{\mathrm{6}} ^{\mathrm{14}} }{\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{3003}}{\mathrm{4}} \\ $$

Commented by jagoll last updated on 23/Feb/20

thank you mister

$${thank}\:{you}\:{mister} \\ $$

Answered by TANMAY PANACEA last updated on 23/Feb/20

(2x+1)^6 ×(((4x^2 +4x+1)/4))^4 (((2x+1)^6 )/4^4 )×{(2x+1)^2 }^4 (((2x+1)^(14) )/4^4 ) now let r+1 th term contains x^6 ((14_C_r )/4^4 )×(2x)^(14−r) ×1^r so x^(14−r) =x^6 →r=8 14c_8 ×(1/4^4 )×2^6 ×x^6

$$\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{6}} ×\left(\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} \\ $$$$\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{6}} }{\mathrm{4}^{\mathrm{4}} }×\left\{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} \right\}^{\mathrm{4}} \\ $$$$\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{14}} }{\mathrm{4}^{\mathrm{4}} } \\ $$$${now}\:{let}\:{r}+\mathrm{1}\:{th}\:{term}\:{contains}\:{x}^{\mathrm{6}} \\ $$$$\frac{\mathrm{14}_{{C}_{{r}} } }{\mathrm{4}^{\mathrm{4}} }×\left(\mathrm{2}{x}\right)^{\mathrm{14}−{r}} ×\mathrm{1}^{{r}} \\ $$$${so}\:{x}^{\mathrm{14}−{r}} ={x}^{\mathrm{6}} \rightarrow{r}=\mathrm{8} \\ $$$$\mathrm{14}{c}_{\mathrm{8}} ×\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{4}} }×\mathrm{2}^{\mathrm{6}} ×{x}^{\mathrm{6}} \\ $$

Commented by jagoll last updated on 23/Feb/20

((14×13×12×11×9×8)/(6×5×4×3×2×1)) × 64

$$\frac{\mathrm{14}×\mathrm{13}×\mathrm{12}×\mathrm{11}×\mathrm{9}×\mathrm{8}}{\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}\:×\:\mathrm{64}\: \\ $$

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