= Solution
A <state on a C-star algebra> is a <positive linear functional> of <norm> one. A <cyclic representation of a C-star algebra> is a representation of a <C-star algebra> $\pi:A\to B(H)$ with a vector $\xi$ whose orbit $\pi(A)\xi$ has dense linear span in $H$. On $c$, pointwise multiplication and coordinatewise <complex conjugation> preserve convergence; the constant sequence one is the unit. The <supremum norm> is submultiplicative and obeys $\|\alpha^*\alpha\|_\infty=\|\alpha\|_\infty^2$. Completeness is given, so \b[$c$ is a commutative unital <C-star algebra>].
If $F=\phi(\beta)$ is a state, testing its positivity on the coordinate projections gives $\beta_{j+1}\geq0$. To recover the independent coefficient at infinity, let $q_N$ have its first $N$ coordinates zero and all later ones one. This is a positive element of $c$, and $F(q_N)=\beta_1+\sum_{j>N}\beta_{j+1}\geq0$. Passing to the limit gives $\beta_1\geq0$. By the given isometry, $\|F\|=\|\beta\|_1=\sum_j\beta_j$, which is one. Conversely these nonnegative unit-sum coefficients make $F(\alpha^*\alpha)$ a sum of nonnegative weighted squares, and the isometry gives <norm> one. Thus \b[the stated coefficient conditions are necessary and sufficient].
For the specified state, all other coefficients are zero because they are nonnegative and the first three already sum to one. Its GNS null ideal is
$$
N_x=\{\alpha:x(\alpha^*\alpha)=0\}=\{\alpha:\alpha_\infty=\alpha_1=\alpha_2=0\}.
$$
Evaluation at infinity and the first two coordinates identifies $c/N_x$ with $\mathbb C^3$, with <inner product> $\tfrac12u_\infty\overline{v_\infty}+\tfrac14u_1\overline{v_1}+\tfrac14u_2\overline{v_2}$. Left multiplication induces the diagonal representation; the class of the unit is $(1,1,1)$. In the equivalent standard <inner product> the resulting GNS triple is
$$
\boxed{H_x=\mathbb C^3,\quad\pi_x(\alpha)=\operatorname{diag}(\alpha_\infty,\alpha_1,\alpha_2),\quad\xi_x=(1/\sqrt2,1/2,1/2).}
$$
Its coefficient is exactly $\tfrac12\alpha_\infty+\tfrac14\alpha_1+\tfrac14\alpha_2=F(\alpha)$. The three evaluations can be assigned arbitrary independent values by a convergent sequence, and all three entries of $\xi_x$ are nonzero. Therefore its orbit spans $\mathbb C^3$, verifying \b[cyclicity and the required state identity] directly.
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