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

jagdish singh singh
173 Points
8 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
8 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
8 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