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Question Number 57863 by mr W last updated on 13/Apr/19

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

Commented by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

(x+x^2 +x^3 +x^4 +x^5 +x^6 )^3 =x^3 (1+x+x^2 +x^3 +x^4 +x^5 )^3 =x^3 ×(((1−x^6 )/(1−x)))^3 =x^3 ×(1−x^6 )^3 ×(1−x)^(−3) =x^3 (1−3x^6 +3x^(12) −x^(18) )(1+3x+6x^2 +10x^3 +..+(((r+1)(r+2))/(1×2))x^r +..) coefficiejt of x →in x^3 is 1→ (k=3) coefficient of x→x^4 →3x^4 is 3→ (k=4) coeeficien of x^ →x^5 →x^3 ×6x^2 is6 → (k=5) coefficient of x→x^6 →x^3 ×10x^3 is10→ (k=6) coefficient of x→x^7 →x^3 ×(((4+1)(4+2))/(1×2))x^4 is 15→(k=7) coefficient of x→x^8 →x^3 ×(((5+1)(5+2))/(1×2))x^5 is21( k=8) wait sir... require d ans=((1+3+6+10+15+21)/(6×6×6))=((56)/(6×6×6))=((14)/(9×6))=(7/(27))

$$\left({x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} +{x}^{\mathrm{6}} \right)^{\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} +{x}^{\mathrm{4}} +{x}^{\mathrm{5}} \right)^{\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} ×\left(\frac{\mathrm{1}−{x}^{\mathrm{6}} }{\mathrm{1}−{x}}\right)^{\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} ×\left(\mathrm{1}−{x}^{\mathrm{6}} \right)^{\mathrm{3}} ×\left(\mathrm{1}−{x}\right)^{−\mathrm{3}} \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{1}−\mathrm{3}{x}^{\mathrm{6}} +\mathrm{3}{x}^{\mathrm{12}} −{x}^{\mathrm{18}} \right)\left(\mathrm{1}+\mathrm{3}{x}+\mathrm{6}{x}^{\mathrm{2}} +\mathrm{10}{x}^{\mathrm{3}} +..+\frac{\left({r}+\mathrm{1}\right)\left({r}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{{r}} +..\right) \\ $$$${coefficiejt}\:{of}\:{x}\:\rightarrow{in}\:{x}^{\mathrm{3}} \:{is}\:\mathrm{1}\rightarrow\:\:\:\:\:\:\left({k}=\mathrm{3}\right) \\ $$$${coefficient}\:{of}\:{x}\rightarrow{x}^{\mathrm{4}} \rightarrow\mathrm{3}{x}^{\mathrm{4}} \:{is}\:\mathrm{3}\rightarrow\:\:\:\:\left({k}=\mathrm{4}\right) \\ $$$${coeeficien}\:{of}\:{x}^{} \rightarrow{x}^{\mathrm{5}} \rightarrow{x}^{\mathrm{3}} ×\mathrm{6}{x}^{\mathrm{2}} \:{is}\mathrm{6}\:\:\rightarrow\:\:\:\left({k}=\mathrm{5}\right) \\ $$$${coefficient}\:{of}\:{x}\rightarrow{x}^{\mathrm{6}} \rightarrow{x}^{\mathrm{3}} ×\mathrm{10}{x}^{\mathrm{3}} \:{is}\mathrm{10}\rightarrow\:\:\:\left({k}=\mathrm{6}\right) \\ $$$${coefficient}\:{of}\:{x}\rightarrow{x}^{\mathrm{7}} \rightarrow{x}^{\mathrm{3}} ×\frac{\left(\mathrm{4}+\mathrm{1}\right)\left(\mathrm{4}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{\mathrm{4}} {is}\:\mathrm{15}\rightarrow\left({k}=\mathrm{7}\right) \\ $$$${coefficient}\:{of}\:{x}\rightarrow{x}^{\mathrm{8}} \rightarrow{x}^{\mathrm{3}} ×\frac{\left(\mathrm{5}+\mathrm{1}\right)\left(\mathrm{5}+\mathrm{2}\right)}{\mathrm{1}×\mathrm{2}}{x}^{\mathrm{5}} \:{is}\mathrm{21}\left(\:{k}=\mathrm{8}\right) \\ $$$${wait}\:{sir}... \\ $$$${require}\:{d}\:{ans}=\frac{\mathrm{1}+\mathrm{3}+\mathrm{6}+\mathrm{10}+\mathrm{15}+\mathrm{21}}{\mathrm{6}×\mathrm{6}×\mathrm{6}}=\frac{\mathrm{56}}{\mathrm{6}×\mathrm{6}×\mathrm{6}}=\frac{\mathrm{14}}{\mathrm{9}×\mathrm{6}}=\frac{\mathrm{7}}{\mathrm{27}} \\ $$$$ \\ $$

Commented by mr W last updated on 14/Apr/19

thank you very much sir! your book is very interesting.

$${thank}\:{you}\:{very}\:{much}\:{sir}! \\ $$$${your}\:{book}\:{is}\:{very}\:{interesting}. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 14/Apr/19

we are student ...we learn everyday...i have learnt from higher algebra Hall and knight...

$${we}\:{are}\:{student}\:...{we}\:{learn}\:{everyday}...{i}\:{have}\:{learnt}\:{from} \\ $$$${higher}\:{algebra}\:{Hall}\:{and}\:{knight}... \\ $$

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