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find the coefficient of x^4 in the expansion of (1+x-2x^2)^7

find the coefficient of x^4 in the expansion of (1+x-2x^2)^7

Grade:11

3 Answers

jagdish singh singh
173 Points
7 years ago
\hspace{-0.7 cm.}$ Let $f(x)=(1+x-2x^2)^7 = a_{0}+a_{1}x+a_{2}x^2+a_{3}x^3+a_{4}x^4+...........$\\\\\\Then $f`(x) = 7(1+x-2x^2)^{6}(1-4x)=a_{1}+2a_{2}x+3a_{3}x^2+4a_{4}x^3...$\\\\\\Now Calculate $f````(x)$ and Put $x=0$ on both side, You get $a_{4}...$
jagdish singh singh
173 Points
7 years ago
\hspace{-0.7 cm.}(1+x-2x^2)^7 = 1+\binom{7}{1}(x-2x^2)+\binom{7}{2}(x-2x^2)^2+$\\\\ $+\binom{7}{3}(x-2x^2)^3+\binom{7}{4}(x-2x^2)^4+........$\\\\\\$ = 1+\binom{7}{2}(x^2+4x^4)+\binom{7}{3}\left[x^3-(2x^2)^3-3x\cdot 2x^2(x-2x^2)\right]+$\\\\\\$+\binom{7}{4}\left[\binom{4}{0}x^4-\binom{4}{1}x^3\cdot (2x^2)+\binom{4}{2}x^2(2x)^2-\binom{4}{1}x(2x)^3+\binom{4}{4}(2x)^4\right]$
 
\hspace{-0.7 cm.}$ So Coeff. of $x^4$ in $(1+x-2x^2)^7 = \binom{7}{2}\times 4+\binom{7}{3}\times (-3\times 2)+$\\\\\\$+\binom{7}{4}+\binom{7}{4}\cdot \binom{4}{2}\times 2^2-\binom{7}{4}\cdot \binom{4}{3}\cdot 2^3+\binom{7}{4}\cdot (2)^4$
RAMCHANDRARAO
159 Points
7 years ago
 IN THE EXPANSION (1+X-2X2 )7 THE COEFICIENT OF X4 IS 4*7C2+7C3*6+7C4  IN THIS EXPANSION YOU  HAVE TO THINK CAREFULLY

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