You can summarize the above networks with 172.1.4.0/22. So how it's done, lets first convert the first two networks to binary (because they have the same netmask of /25):

10101100.00000001.00000100.**0**XXXXXXX /25

10101100.00000001.00000100.**1**XXXXXXX /25

X's here means the host bits which can be 0 or 1. You can see that the first bit of the 4th octet is 0 in the first network and 1 in the second network. This means that those two networks can be summarized by one network looking like this:

10101100.00000001.00000100.**X**XXXXXXX /24

In other words turn the changing bit into an X and subtract 1 from the subnet mask (i.e 25 - 1 = 24). Back to decimal gives us:

172.1.4.0/24

Now we have the following networks:

172.1.4.0/24

172.1.5.0/24

172.1.6.0/24

172.1.7.0/24

Again since all have the same subnet mask may be we can summarize further. So again we will find the binary of all:

10101100.00000001.000001**00**.XXXXXXXX /24

10101100.00000001.000001**01**.XXXXXXXX /24

10101100.00000001.000001**10**.XXXXXXXX /24

10101100.00000001.000001**11**.XXXXXXXX /24

Notice how the two last bits of the 3rd octet are changing. All combinations are there. 00, 01, 10 and 11. There are no more combinations for 2 bits. This means that those four networks can be summerized by one network looking like this:

10101100.00000001.000001**XX**.XXXXXXXX /22

In other words, turn the two changing bits into XX and subtract 2 from the subnet mask (i.e 24 - 2 = 22). Turning it back to decimal we get:

172.1.4.0/22

You can do this for 3,4,5.....etc changing bits. The thing that you have to make sure of is that **ALL COMBINATIONS ARE PRESENT IN THE CHANGING BITS**. Ofcourse, the networks that you want to summarize need to have the SAME subnet mask too.

This looks too lengthy only because I'm trying to explain it. You can do it much faster once you get used to it.