[Physics] Ricci tensor for a 3-sphere without Math packets

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Let's have the metric for a 3-sphere:
$$dl^{2} = R^{2}\left(d\psi ^{2} + sin^{2}(\psi )(d \theta ^{2} + sin^{2}(\theta ) d \varphi^{2})\right).$$
I tried to calculate Riemann or Ricci tensor's components, but I got problems with it.

In the beginning, I got an expressions for Christoffel's symbols:
$$\Gamma^{i}_{ii} = \frac{1}{2}g^{ii}\partial_{i}g_{ii} = 0,$$

$$\Gamma^{i}_{ji} = \frac{1}{2}g^{ii}\partial_{j}g_{ii},$$

$$\Gamma^{k}_{ll} = -\frac{1}{2}g^{kk}\partial_{k}g_{ll},$$

$$\Gamma^{k}_{lj} = \Gamma^{k}_{lk}\delta^{k}_{j} + \Gamma^{k}_{jk}\delta^{k}_{l} + \Gamma^{k}_{jj}\delta^{j}_{l} = 0.$$

The Ricci curvature must be
$$R_{lj}=\frac{2}{R^{2}}g_{lj}.$$
But when I use definition of Ricci tensor,

$$R_{lj}^{(3)} = \partial_{k}\Gamma^{k}_{lj} – \partial_{l}\Gamma^{\lambda}_{j \lambda} + \Gamma^{k}_{j l}\Gamma^{\sigma}_{k \sigma} – \Gamma^{k }_{l \sigma}\Gamma^{\sigma}_{jk},$$
I can't associate an expression (if I didn't make the mistakes)

$$R_{lj}^{(3)} = \partial_{j}\Gamma^{j}_{lj} + \partial_{l}\Gamma^{l}_{jl} + \partial_{k}\Gamma^{k}_{ll}\delta^{l}_{j} – \partial_{l}\Gamma^{k}_{jk} – \Gamma^{k}_{jk}\Gamma^{j}_{lj} + \Gamma^{k}_{lk}\Gamma^{l}_{jl} + \Gamma^{\sigma}_{k \sigma}\Gamma^{k}_{ll}\delta^{l}_{j} – \Gamma^{k}_{jk}\Gamma^{k}_{lk} – \Gamma^{l}_{jl}\Gamma^{j}_{lj} – \Gamma^{l}_{kl}\Gamma^{k}_{ll}\delta^{l}_{j} – \Gamma^{j}_{ll}\Gamma^{l}_{jj} – \Gamma^{k}_{ll}\Gamma^{l}_{kl} =$$

$$= \partial_{j}\Gamma^{j}_{lj} + \partial_{l}\Gamma^{l}_{jl} + \partial_{k}\Gamma^{k}_{ll}\delta^{l}_{j} – \partial_{l}\Gamma^{k}_{jk} – \Gamma^{k}_{jk}\Gamma^{j}_{lj} + \Gamma^{k}_{lk}\Gamma^{l}_{jl} + \Gamma^{\sigma}_{k \sigma}\Gamma^{k}_{ll}\delta^{l}_{j} – \Gamma^{k}_{jk}\Gamma^{k}_{lk} – \Gamma^{l}_{jl}\Gamma^{j}_{lj} – 2\Gamma^{l}_{kl}\Gamma^{k}_{ll}\delta^{l}_{j} – \Gamma^{j}_{ll}\Gamma^{l}_{jj},$$
where there is a summation only on $k, \sigma$, with an expression for the metric tensor.

Maybe, there are some hints, which can help?

Take the coordinates $x_i,\,x=\sqrt{x_i x_i}$ to be (I write $r$ so it cant be confused with $R$ the contraction of Ricci tensor): $$\psi=x/r\\ x_1=r\psi\sin\theta\cos\phi\\ x_2=r\psi\sin\theta\sin\phi\\ x_3=r\psi\cos\theta$$ Compare it with usual spherical coordinates in $\mathbb{R}^3$, we know: $$dx_idx_i=r^2d\psi^2+r^2\psi^2(d\theta^2+\sin^2\theta d\phi^2)$$ So we can rewrite your metric as $$ds^2=\frac{\sin^2\psi}{\psi^2}dx_i dx_i+r^2d\psi^2\left(1-\frac{\sin^2\psi}{\psi^2}\right)=\\ =r^2\frac{\sin^2(x/r)}{x^2}dx_i dx_i+\frac{(x_idx_i)^2}{x^2}\left(1-r^2\frac{\sin^2(x/r)}{x^2}\right)$$
Now, if we believe that $x_i$ are normal, then we have: $$g_{ik}(x)=\delta_{ik}-\frac{1}{3}R_{iakb}x^ax^b+...$$ So you can just expand the metric and find the symmetrized form $\frac{1}{2}\left(R_{iakb}+R_{ibka}\right)$ of the Riemann tensor at the origin that is sufficient to contract it to the Ricci tensor. The good point is that it is the same level of complexity (in fact, the 'normal' metric of the same form) in any dimension. As soon as you will find $R_{ik}\sim g_{ik}$ in this coordinate system, it holds in any coordinate system. And, due to the symmetry of the sphere, it also holds at every point.