Question Number 54024 by sandip123 last updated on 28/Jan/19
$$\mathrm{The}\:\mathrm{coefficient}\:\mathrm{of}\:\mathrm{x}^{\mathrm{4}} \:\mathrm{in}\:\left(\mathrm{2}−\mathrm{4x}+\mathrm{3x}^{\mathrm{2}} \right)^{−\mathrm{2}} \mathrm{i}\:\mathrm{s}\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19
![(2−4x+3x^2 )^(−2) ={2−(4x−3x^2 )}^(−2) =2^(−2) {1−(((4x−3x^2 )/2))}^(−2) now using formula (1−y)^(−2) =1+2y+3y^2 +4y^3 +...+(r+1)x^r +... =2^(−2) [1+2(((4x−3x^2 )/2))+3(((4x−3x^2 )/2))^2 +4(((4x−3x^2 )/2))^3 +5(((4x−3x^2 )/2))^4 +... =2^(−2) [1+4x−3x^2 +(3/4)(4x^2 −24x^3 +9x^4 )+(4/8)(64x^3 −3×16x^2 ×3x^2 +3×4x×9x^4 −27x^6 )+(5/2^4 )(4^4 x^4 +others ignored) ignoring red marked... =2^(−2) [(3/4)×9x^4 +(4/8)×−9×16×x^4 +(5/2^4 )×4^4 x^4 ] =x^4 [((3×9)/(4×2^2 ))−((4×9×16)/(2^2 ×8))+((5×4^4 )/(2^4 ×2^2 ))] =x^4 [((27)/(16))−((18)/1)+((5×2^8 )/2^6 )] =x^4 [((27)/(16))−((18)/1)+((5×2^2 )/1)] =x^4 [((27)/(26))+2] so coefficient=((79)/(26)) pls check steps...](https://www.tinkutara.com/question/Q54030.png)
$$\left(\mathrm{2}−\mathrm{4}{x}+\mathrm{3}{x}^{\mathrm{2}} \right)^{−\mathrm{2}} \\ $$$$=\left\{\mathrm{2}−\left(\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} \right)\right\}^{−\mathrm{2}} \\ $$$$=\mathrm{2}^{−\mathrm{2}} \left\{\mathrm{1}−\left(\frac{\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)\right\}^{−\mathrm{2}} \\ $$$${now}\:{using}\:{formula} \\ $$$$\left(\mathrm{1}−{y}\right)^{−\mathrm{2}} =\mathrm{1}+\mathrm{2}{y}+\mathrm{3}{y}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{3}} +…+\left({r}+\mathrm{1}\right){x}^{{r}} +… \\ $$$$=\mathrm{2}^{−\mathrm{2}} \left[\mathrm{1}+\mathrm{2}\left(\frac{\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)+\mathrm{3}\left(\frac{\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{3}} +\mathrm{5}\left(\frac{\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{4}} +…\right. \\ $$$$=\mathrm{2}^{−\mathrm{2}} \left[\mathrm{1}+\mathrm{4}{x}−\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{24}{x}^{\mathrm{3}} +\mathrm{9}{x}^{\mathrm{4}} \right)+\frac{\mathrm{4}}{\mathrm{8}}\left(\mathrm{64}{x}^{\mathrm{3}} −\mathrm{3}×\mathrm{16}{x}^{\mathrm{2}} ×\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}{x}×\mathrm{9}{x}^{\mathrm{4}} −\mathrm{27}{x}^{\mathrm{6}} \right)+\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{4}} }\left(\mathrm{4}^{\mathrm{4}} {x}^{\mathrm{4}} +{others}\:{ignored}\right)\right. \\ $$$${ignoring}\:\:{red}\:{marked}… \\ $$$$=\mathrm{2}^{−\mathrm{2}} \left[\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{9}{x}^{\mathrm{4}} +\frac{\mathrm{4}}{\mathrm{8}}×−\mathrm{9}×\mathrm{16}×{x}^{\mathrm{4}} +\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{4}} }×\mathrm{4}^{\mathrm{4}} {x}^{\mathrm{4}} \right] \\ $$$$={x}^{\mathrm{4}} \left[\frac{\mathrm{3}×\mathrm{9}}{\mathrm{4}×\mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{4}×\mathrm{9}×\mathrm{16}}{\mathrm{2}^{\mathrm{2}} ×\mathrm{8}}+\frac{\mathrm{5}×\mathrm{4}^{\mathrm{4}} }{\mathrm{2}^{\mathrm{4}} ×\mathrm{2}^{\mathrm{2}} }\right] \\ $$$$={x}^{\mathrm{4}} \left[\frac{\mathrm{27}}{\mathrm{16}}−\frac{\mathrm{18}}{\mathrm{1}}+\frac{\mathrm{5}×\mathrm{2}^{\mathrm{8}} }{\mathrm{2}^{\mathrm{6}} }\right] \\ $$$$={x}^{\mathrm{4}} \left[\frac{\mathrm{27}}{\mathrm{16}}−\frac{\mathrm{18}}{\mathrm{1}}+\frac{\mathrm{5}×\mathrm{2}^{\mathrm{2}} }{\mathrm{1}}\right] \\ $$$$={x}^{\mathrm{4}} \left[\frac{\mathrm{27}}{\mathrm{26}}+\mathrm{2}\right]\:\:{so}\:{coefficient}=\frac{\mathrm{79}}{\mathrm{26}} \\ $$$${pls}\:{check}\:{steps}… \\ $$
Commented by sandip123 last updated on 01/Feb/19
$$\mathrm{you}\:\mathrm{will}\:\mathrm{get}\:\mathrm{right}\:\mathrm{answer}=\mathrm{59}/\mathrm{16}\: \\ $$$$\mathrm{error}\:\mathrm{is}\:\mathrm{only}\:\mathrm{in}\:\mathrm{last}\:\mathrm{line}.\:\:\:\:\mathrm{thanks}! \\ $$