The 10^th term will be

19 x 1 + 17 x 2 + 15 x 3 + 13 x 4 + 11 x 5 + 9 x 6 + 7 x 7 + 5 x 8 + 3 x 9 + 1 x 10 = 385

The 100^th term can be calculated in the same way (if you have the time and the determination), or you might be able to write a Python program to do the calculations for you.

Alternatively, you could try to find a general formula for the n^th term using a difference method.

Sequence: 1, 5, 14, 30, 55…

First difference 4, 9, 16, 25… (this looks familiar!)

Second difference 5, 7, 9, …. (so does this!)

Third difference 2, 2, … (this is constant )

When the third difference is constant, we know the formula for the n^thterm is of the form

an³ + bn² + cn + d

By substituting n = 1, 2, 3, 4 we get four equations to solve simultaneously

a + b + c + d = 1

8a + 4b + 2c + d = 1

27a + 9b + 3c + d = 1

64a + 16b + 4c + d = 1

This is long (but not too difficult) and produces the solution a=1/3, b = 1/2, c = 1/6, d = 0.

So the n^th term is n³/3+ n²/2+ n/6 = (2n³ + 3n² + n)/6 = n(n + 1)(2n + 1)/6

Therefore the 100^th term is 100(100 + 1)(2x100 + 1)/6 = 100 x 101 x 201/6 = 50 x 101 x 67 = 338,35

Now, n(n + 1)(2n + 1)/6 is the formula for the sum of the first n square numbers (there are lots of places to look this up online, if you want to know where it comes from), and there are two reasons why this should not be a surprise.

The first difference above shows us that the totals for each square are obtained by adding the next square number, so the n^th term is 1 + 4 + 9 + 16 + 25 + … + n²
Each diagram can be split up to show why there are square numbers involved, for example:

1	1	1	1		1	1	1	1
1	2	2	2		1	1	1	1		1	1	1
1	2	3	3	=	1	1	1	1	+	1	1	1	+	1	1	+
1	2	3	4		1	1	1	1		1	1	1		1	1		1

30 = 16 + 9 + 4 + 1

The total for the fourth square in the family can be calculated as 1 x 7 + 2 x 5 + 3 x 3 + 4 x 1, but this must be the same as 16 + 9 + 4 + 1.

Challenge 327: Nic's Squares