Interpolated FIR filter I am confused about this Q&amp;A :  Interpolation by factor of 2

If my input signal 
$$
x[n]= x0,x1,x2,x3
$$
then according to the threads explaination my $v[n]$ will be 

$$
v[n]=x0,0,x1,0,x2,0,x3,0
$$

$x[n] \rightarrow$ zeropad $\rightarrow v[n]$

$v[n]\rightarrow$ LPF $\rightarrow y[m]$

My $y[m]$ is the output of LPF which removes aliases. So basically LPF transfer function is given as 

$$
y[m] = v[n]*h[n]
$$

i,e. $y[m]=\displaystyle\sum_{n=0}^{\infty}h(m-n)v(n)$

So here $h[m]$ is my filter coefficients and  $v[n]$ is my zero padded input signal...
so, if I multiply zero with my filter coefficients, then my output will also be zero...

If I send 10 input samples, after zero padding I will get 20 samples, out of which 10 will be zero....

So what is the use of getting zero as my output? Correct me if am wrong (I know I am wrong, but please give me some idea how zero padding &amp; LPF is helping in this regard).

For FIR filter: $y(n)=h(0)x(n)+h(1)x(n−1)+....+h(N−1)x(n−N+1)$,
 where $x(n)$ is the input signal,
 $y(n)$ is the output signal, 
$h(n)$ is the impulse response, 
and $N$ is the filter length. 

For simplicity, I am taking order $N=4$,and some random numbers

    (time) n --&gt; [ 0  1  2  3] 
         x[n]--&gt; [ 2  4  6  8] 
         h[n]--&gt; [.1 .2 .3 .4] 


what is $y[2.5]$??? 

                  n--&gt;[0 0.5 1 1.5 2 2.5 3 3.5]  
               v[n]--&gt;[2  0  4  0  6  0  8  0 ] (after zero padding)


Lets calculate for $y[2.5]=h(0)x(2.5)+h(1)x(1.5)+h(2)x(0.5)+h(3)x(-0.5)$ 

so here $x(2.5)=x(1.5)=x(0.5)=0$

Please answer this to clear my confusion???
The interpolation lowpass filter computes a weighted sum of input samples, which results in the zero input samples being interpolated using the non-zero samples of the input signal. The $*$ sign is NOT multiplication but convolution!

EDIT: I'm adding a simple example to clear things up a bit. Let's consider the impulse response $h = [0.5, 1, 0.5]$ and a zero-padded input signal
$x=[1,0,3,0,5,0,3,0,1]$. The time index is denoted by $n$, and $n=0$ corresponds to the left-most values of $h$ and $x$. The output $y$ is the convolution of $(x*h)(n)$:

$$y(0) = h(0)x(0) = 0.5\\
y(1) = h(0)x(1) + h(1)x(0) = 1\\
y(2) = h(0)x(2) + h(1)x(1) + h(2)x(0) = 2\\
y(3) = h(0)x(3) + h(1)x(2) + h(2)x(1) = 3\\
y(4) = h(0)x(4) + h(1)x(3) + h(2)x(2) = 4\\
y(5) = h(0)x(5) + h(1)x(4) + h(2)x(3) = 5\\\vdots$$

This impulse response obviously performs linear interpolation of the zero-padded input signal.