Understanding subscripts in first order conditions of dynamic optimization problems Suppose we have a simple maximization problem as described in Equation 1.1 here or here. This leads us to the Lagrangian Equation 1.3: $$\begin{align*}\mathcal{L} &amp;= \sum_{t=1}^\infty \beta^{t-1}\left\{u(c_t) + \lambda_t \left[ f(k_t) + (1 - \delta)k_t - c_t - k_{t+1}\right]\right\} \\ &amp;= \sum_{t=1}^\infty \left[\beta^{t-1} u(c_t) - \beta^{t-1}\lambda_t c_t + \beta^{t-1} \lambda_t f(\mathbf{k_t}) + \beta^{t-1}\lambda_t(1-\delta)\mathbf{k_t}  - \beta^{t-1}\lambda_t \mathbf{k_{t+1}}\right]  \end{align*}
$$

When we derive the first order condition with respect to $k_{t+1}$, which is:
$$\frac{\partial \mathcal{L} (\cdot)}{\partial k_{t+1}} = 0 :   \beta \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta \lambda_{t+1} (1 - \delta) -\lambda_t=0$$

why do we use the subscript $\phantom{.}_{t+1}$ in  $\lambda_{t+1}$ and why does $\beta^{t-1}$ becomes $\beta$? I cannot understand how the first two terms are combined with the last one ($-\lambda_t$). 

The relevant terms (with $k$) of the Lagrangian in period $\phantom{.}_{t+1}$ are: $$ \beta^{(t+1)-1} \lambda_{t+1} f(k_{t+1}) +  \beta^{(t+1)-1} \lambda_{t+1} k_{t+1} (1 - \delta) - \beta^{(t+1)-1} \lambda_{t+1} k_{(t+1)+1}$$ so for this part of the sum we do not "care" about the last term when we take the derivative with respect to $k_{t+1}$. So for this period this part of the sum is $$\frac{\partial \mathcal{L}_{t+1}}{\partial k_{t+1}} = \beta^{t} \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta^{t} \lambda_{t+1}  (1 - \delta)$$ 

The relevant terms (with $k$)  of the Lagrangian in period $\phantom{.}_{t}$ are:
$$\beta^{t-1} \lambda_t f({k_t}) + \beta^{t-1}\lambda_t(1-\delta){k_t}  - \beta^{t-1}\lambda_t k_{t+1}$$
so for this period the part of the sum is $$\frac{\partial \mathcal{L}_t}{\partial k_{t+1}} =  - \beta^{t-1}\lambda_t$$
Now the First Order Condition with respect to $k_{t+1}$ should be:  $$\frac{\partial \mathcal{L}_{t+1}}{\partial k_{t+1}} +  \frac{\partial \mathcal{L}_t}{\partial k_{t+1}} = \beta^{t} \lambda_{t+1} \frac{\partial f(k_{t+1})}{\partial k_{t+1}} +  \beta^{t} \lambda_{t+1}  (1 - \delta) - \beta^{t-1}\lambda_t = 0$$right?
In an intertemporal maximization problem, we seek to find the optimal sequence  of the control and the state variables. It is the recursive nature of the problem that permits us to consider a "typical" point in time and just one condition per variable.  

For each such problem, we need to find out (carefully) in how many distinct periods a specific realization of a variable appears. To do this properly we should distinguish between the "absolute" index, and a "running" index. In the formulation of the Lagrangean as appears in the question, this is not done (and it is usual practice not to, but it may become confusing).  

So I would use the $t$ symbol as the absolute index (to arrive at same-looking first-order conditions), and some other symbol for the running index, say

$$\mathcal{L_t} = \sum_{j=0}^\infty \beta^{j}\left\{u(c_{t+j}) + \lambda_{t+j} \left[ f(k_{t+j}) + (1 - \delta)k_{t+j} - c_{t+j} - k_{t+j+1}\right]\right\} $$

Note that $t$ no longer affects  the discount factor $\beta$, and this is because the discount factor has to do with looking at the future, which is represented by the index $j$. Also, note that $j$ starts at zero, indicating that the first period is the $t$ period.  

Written this way, the Lagrangean says "we are at some point in time indicated by $t$ (that can take the value zero or whatever positive value), and we are looking forward period by period counted by index $j$".

For any given $j$ we have

$$\mathcal{L}_t =...+ \beta^{j}\Big\{u(c_{t+j}) + \lambda_{t+j} \left[ f(k_{t+j}) + (1 - \delta)k_{t+j} - c_{t+j} - k_{t+j+1}\right]\Big\} + \beta^{j+1}\Big\{u(c_{t+j+1}) + \lambda_{t+j+1} \left[ f(k_{t+j+1}) + (1 - \delta)k_{t+j+1} - c_{t+j+1} - k_{t+j+2}\right]\Big\} + ...$$

Pondering this, we realize that the variable $k_{t+j+1}$ will appear in only two consecutive periods, and so the first order condition for a "typical" element of the sequence $\{k_{t+j}\}_{j=0}^{\infty}$ can be exrpessed by differentiating only these two periods with respect to $k_{t+j+1}$. Doing so we get

$$\frac {\partial \mathcal{L}_t}{\partial k_{t+j+1}} =  -\beta^{j} \lambda_{t+j}  + \beta^{j+1}\Big\{ \lambda_{t+j+1} \left[ f'(k_{t+j+1}) + (1 - \delta)\right]\Big\} $$

Take common factors (which will simplify the discount factor) and set equal to zero

$$\frac {\partial \mathcal{L}_t}{\partial k_{t+j+1}} = \beta^{j} \Big[-\lambda_{t+j}  + \beta\Big\{ \lambda_{t+j+1} \left[ f'(k_{t+j+1}) + (1 - \delta)\right]\Big\}\Big] = 0$$

To lighten the indexing burden, we can express this for $j=0$, to obtain

$$\frac {\partial \mathcal{L}_t}{\partial k_{t+1}} = 0 \implies -\lambda_{t}  + \beta\Big\{ \lambda_{t+1} \left[ f'(k_{t+1}) + (1 - \delta)\right]\Big\} = 0$$