Let K be the number of rational terms in the | vedclass

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(C) The general term in the expansion of $(\sqrt{2}+\sqrt[3]{3})^{6144}$ is $T_{r+1} = \binom{6144}{r} (2^{1/2})^{6144-r} (3^{1/3})^r = \binom{6144}{r

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-$2$

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(C) The general term in the expansion of $(\sqrt{2}+\sqrt[3]{3})^{6144}$ is $T_{r+1} = \binom{6144}{r} (2^{1/2})^{6144-r} (3^{1/3})^r = \binom{6144}{r} 2^{(6144-r)/2} 3^{r/3}$.For the term to be rational,$(6144-r)/2$ and $r/3$ must be integers.This implies $r$ must be a multiple of $3$ and $6144-r$ must be even (which is true for any even $r$).Thus,$r$ must be a multiple of $6$. Let $r = 6k$,where $0 \le 6k \le 6144$,so $0 \le k \le 1024$.The number of values for $k$ is $1024 - 0 + 1 = 1025$. Thus,$K = 1025$.Now,consider the product $f(x) = \frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})}$.Using the identity $(1-x)(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16}) = 1-x^{32}$,we have $f(x) = \frac{1-x}{1-x^{32}} = (1-x)(1+x^{32}+x^{64}+\dots)$.This means the coefficient $\alpha_P$ is $1$ if $P$ is a multiple of $32$,$-1$ if $P-1$ is a multiple of $32$,and $0$ otherwise.For $K = 1025$,$1025 = 32 	imes 32 + 1$. Thus $\alpha_{1025} = -1$ (since $1025-1 = 1024 = 32 	imes 32$).For $K+1 = 1026$,$\alpha_{1026} = 0$.For $K-1 = 1024$,$\alpha_{1024} = 1$ (since $1024 = 32 	imes 32$).Therefore,$\alpha_{K}-\alpha_{K+1}-\alpha_{K-1} = -1 - 0 - 1 = -2$.

Let $K$ be the number of rational terms in the expansion of $(\sqrt{2}+\sqrt[3]{3})^{6144}$. If the coefficient of $x^{P} \quad(P \in N)$ in the expansion of $\frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})}$ is $\alpha_{P}$,then $\alpha_{K}-\alpha_{K+1}-\alpha_{K-1}=$

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