The sum of the binomial coefficients of [2x + | vedclass

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Question

(A) The sum of binomial coefficients of $(a + b)^n$ is $2^n$.Given $2^n = 256 = 2^8$,so $n = 8$.The general term in the expansion of $[2x + x^{-1}]^8$

Vedclass · Accepted Answer

$1120$

Vedclass · Answer

(A) The sum of binomial coefficients of $(a + b)^n$ is $2^n$.Given $2^n = 256 = 2^8$,so $n = 8$.The general term in the expansion of $[2x + x^{-1}]^8$ is $T_{r+1} = \binom{8}{r} (2x)^{8-r} (x^{-1})^r$.$T_{r+1} = \binom{8}{r} 2^{8-r} x^{8-r} x^{-r} = \binom{8}{r} 2^{8-r} x^{8-2r}$.For the constant term,the power of $x$ must be $0$,so $8 - 2r = 0 \Rightarrow r = 4$.The constant term is $\binom{8}{4} 2^{8-4} = \binom{8}{4} 2^4$.$\binom{8}{4} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.Constant term $= 70 	imes 16 = 1120$.

The sum of the binomial coefficients of $[2x + \frac{1}{x}]^n$ is equal to $256$. The constant term in the expansion is

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