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Powers of 11 and powers of 9

Many people, especially younger students, have noticed the pattern of the numbers in the first few rows of Pascal's Triangle follows the pattern of the digits of the powers of 11. 110 = 1 111 = 11 112 = 121 113 = 1331 etc.   This, as you might expect, is no accident; as we know from the binomial theorem, , and if we let x=10 and y=1, this becomes ; for the first 4 rows of Pascal's Triangle, where all the entries are less than 10, the numbers read across exactly like the powers of 11, but once a row has entries with two digits or more we have to be more careful.  For example, row 7 of Pascal's Triangle is 1 7 21 35 35 21 7 1; to get 117, we need to do a little more work.        1 =  1 * 100       70 =  7 * 101     2100 = 21 * 102    35000 = 35 * 103   350000 = 35 * 104  2100000 = 21 * 105  7000000 =  7 * 106 10000000 =  1 * 107 19487171 = 117

It's not quite as obvious that the powers of 9 are hidden in the rows of Pascal's Triangle, but if we let x=10 and y=-1 in the binomial theorem, we get .  Now if we take any row after row zero, we can group the coefficients to make two numbers in the following way. 90 = 1 91 = 10 - 1 = 9 92 = 101 - 20 = 81 93 = 1030 - 301 = 729   Here's the idea of the pattern for row 7; we will color the coefficients alternately red and blue, get both sums then subtract the blue sum from the red. Step 1:  Alternating colors: 1 7 21 35 35 21 7 1 Step 2:  Red sum and blue sum       70 =  7 * 101    35000 = 35 * 103   2100000 = 21 * 105 10000000 =  1 * 107 12135070        1 =  1 * 100     2100 = 21 * 102   350000 = 35 * 104   7000000 =  7 * 106   7352101 Step 3: Red - Blue 12135070 -7352101  4782969 = 97