How do you find the coefficient of $x^5$ in the expansion of $(x-3)^7$ ?

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Answer 1 · 2016-12-10T14:56:57

189

for integers the binomial expansion is

$(x+y)^n=x^n+^nC_1x^(n-1)y+^nC_2n^(n-2)y^2+....^nC_kx^(n-k)y^k+.....y^n$

where $" "^nC_k=(n!)/(k!(n-k)!)$

for $" "(x-3)^7$

the term including $" "x^5$

$" "^7C_2x^5(-3)^2$

$=(7!)/(2!(5!))xx9$

$=((7xx6xxcancel(5!))/(2xxcancel(5!)))xx9$

$=21xx9=189$