Find the value of k such that the coefficient of the x^-1 term in the expansion of (kx+2x)^5 is 720.
Added by Devin G.
Step 1
First, we need to expand (kx+2x)^5 using the binomial theorem: (kx+2x)^5 = C(5,0)(kx)^5 + C(5,1)(kx)^4(2x) + C(5,2)(kx)^3(2x)^2 + C(5,3)(kx)^2(2x)^3 + C(5,4)(kx)(2x)^4 + C(5,5)(2x)^5 where C(n,r) represents the binomial coefficient "n choose r" or "nCr". Show more…
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