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Coefficiants of (2x+y)^4​

1 Answer

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By the binomial theorem,


(2x+y)^4=\displaystyle\sum_(k=0)^4\binom 4k(2x)^(4-k)y^k=\sum_(k=0)^4\binom 4k2^(4-k)x^(4-k)y^k

where


\dbinom nk=(n!)/(k!(n-k)!)

Then the coefficients of the
x^(4-k)y^k terms in the expansion are, in order from
k=0 to
k=4,


\dbinom 402^(4-0)=1\cdot2^4=16


\dbinom412^(4-1)=4\cdot2^3=32


\dbinom422^(4-2)=6\cdot2^2=24


\dbinom432^(4-3)=4\cdot2^1=8


\dbinom442^(4-4)=1\cdot2^0=1

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