How can I find components of a (1,2) tensor? Can someone please help me with the following questions? I am really lost as to what they are asking or how I can go about solving them. I tried my best to analyze the question as much as possible. 
Let $A_{jk}^{i}$ be the components of a $(1,2)$ tensor on $\mathbb{R}^3.$ Suppose at a point $x \in \mathbb{R}^3, A_{jk}^{i} = (i+j)k$ at $x$. 
a. Evaluate $A_{i2}^i$
I know that this is $(i+i)2 = (2i)2 = 4i$ but I don't know what this means or if this is the answer. 
b. Evaluate $A_{3\alpha}^\alpha$
Again, it's the same problem so I have $(\alpha+3)\alpha = \alpha^2 +3\alpha$ 
c. How many components does the original tensor A have?
d. How many components does the tensor A have after we contract on $i=j$?
I think that it's asking for what $A_{ik}^{i}$
And I'm unsure about the last two questions, thanks for any help! 
 A: $A^i_{jk}$ is a mixed tensor of rank 3 with 1 contravariant $i$ index and 2 variant $j,k$ indexes, having in total $n$=3 indexes and defined in a $m$=3 dimensional space. 
a. Given $A^i_{jk}=(i+j)k$, evaluate $A^i_{i2}$
Contracting the first contravariant (upper) and first covariant (lower) terms:
$A^i_{ik}=\sum_{i=1}^3(i+i)k=\sum_{i=1}^{n}2ik=2\frac{n(n+1)}2k$
For a 3 dimensional space, $n=3$ (of course the sum formula is redundant in here):
$A^i_{ik}=12k$
The operation reduces the rank from 3 into rank 3-2=1, a vector.
The second component of this vector, $k=2$ is $A^i_{i2}=24$.
b. Contracting the first contravariant (upper) and second covariant (lower) terms (the contracted index variable is dummy):
$A^i_{ji}=\sum_{i=1}^3(i+j)i=\sum_{i=1}^3 i^2+ij=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}{2}j$
Again, for $n=3$, $A^i_{ji}=20j$.
The operation reduces the rank from 3 into rank 3-2=1, a vector, as expected.
The third component of our new vector $j=3$ is $A^i_{3i}=60$.
c. In a $m$ dimensional space, a scalar will have 1 component (rank 0), a vector will have 3 components (rank 1), a matrix will have 9 components (rank 2), and in our case, the mixed (1,2), 3 rank tensor will have $m^n=3^3=27$ components. In a cartesian view, this is like a matricial cube.
d. An index contraction is the operation of equaling one contravariant (upper) index with a variant (lower) index, hance reducing the rank of the tensor by two, and summing the resulting components. 
In a scalar, there is no significative meaning, just the identity; in a vector, there is just the first component taken; in a matrix, the resulting value is the trace. 
In our case (both previous a. and b. cases without taking a specific component), the operation is understood as a generalization of the trace for the contracted indexes, reducing from rank 3 into rank 1, a vector, as shown, hence in that case the result is a vector with $m^1=3$ components.
