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Question-121889




Question Number 121889 by oustmuchiya@gmail.com last updated on 12/Nov/20
Answered by Dwaipayan Shikari last updated on 12/Nov/20
(1+ax)^b =1+48x+1008x^2 +...  ab=48  a^2 ((b(b−1))/2)=1008  ((a(b−1))/2)=21⇒a=6 ,b=8
$$\left(\mathrm{1}+{ax}\right)^{{b}} =\mathrm{1}+\mathrm{48}{x}+\mathrm{1008}{x}^{\mathrm{2}} +… \\ $$$${ab}=\mathrm{48} \\ $$$${a}^{\mathrm{2}} \frac{{b}\left({b}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1008} \\ $$$$\frac{{a}\left({b}−\mathrm{1}\right)}{\mathrm{2}}=\mathrm{21}\Rightarrow{a}=\mathrm{6}\:,{b}=\mathrm{8} \\ $$
Answered by TANMAY PANACEA last updated on 12/Nov/20
(1+2x+3x^2 )(1+2x)^7   (1+2x+3x^2 ){1+7C_1 (2x)+7C_2 (2x)^2 +7C_3 (2x)^3 +othdr terms}  =1×7C_3 (2x)^3 +2x×7C_2 (2x)^2 +3x^2 ×7C_1 (2x)  =x^3 (7C_3 ×2^3 +2×7C_2 ×2^2 +3×7C_1 ×2)   pls calculate
$$\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{7}} \\ $$$$\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} \right)\left\{\mathrm{1}+\mathrm{7}{C}_{\mathrm{1}} \left(\mathrm{2}{x}\right)+\mathrm{7}{C}_{\mathrm{2}} \left(\mathrm{2}{x}\right)^{\mathrm{2}} +\mathrm{7}{C}_{\mathrm{3}} \left(\mathrm{2}{x}\right)^{\mathrm{3}} +{othdr}\:{terms}\right\} \\ $$$$=\mathrm{1}×\mathrm{7}{C}_{\mathrm{3}} \left(\mathrm{2}{x}\right)^{\mathrm{3}} +\mathrm{2}{x}×\mathrm{7}{C}_{\mathrm{2}} \left(\mathrm{2}{x}\right)^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} ×\mathrm{7}{C}_{\mathrm{1}} \left(\mathrm{2}{x}\right) \\ $$$$={x}^{\mathrm{3}} \left(\mathrm{7}{C}_{\mathrm{3}} ×\mathrm{2}^{\mathrm{3}} +\mathrm{2}×\mathrm{7}{C}_{\mathrm{2}} ×\mathrm{2}^{\mathrm{2}} +\mathrm{3}×\mathrm{7}{C}_{\mathrm{1}} ×\mathrm{2}\right) \\ $$$$\:{pls}\:{calculate} \\ $$

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