A Pattern in the Power Series Coefficients and Pascal’s Triangle

An interesting pattern emerges from evaluating the sum of descending powers that alternate in sign with Pascal’s Triangle coefficients.

Pascal’s Triangle is probably familiar to most students of mathematics. The pattern is shown below.

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

…

We will take the descending powers with alternating signs and multiply them with the numbers in Pascal’s Triangle. This is the pattern we get for the sums.

\(2^x\)

\(3^x - 2^x\)

\(4^x - 2\cdot 3^x + 2^x\)

\(5^x - 3 \cdot 4^x + 3 \cdot 3^x - 2^x \)

\( 6^x - 4 \cdot 5^x + 6 \cdot 4^x - 4 \cdot 3^x + 2^x \)

\( 7^x - 5 \cdot 6^x + 10 \cdot 5^x - 10 \cdot 4^x + 5 \cdot 3^x - 2^x \)

…

When evaluating the sum for integer values of x starting with 0, we get 0 as the result until we get a non-zero answer. Let’s evaluate this sum for x = 0, 1, 2, 3, 4, .... until we get a non-zero answer for 2^x.

\(2^x\)

\( 2^0 = 1 \)

Okay. That was obvious. The first value gives us 1. But let’s continue with the next sum:

\( 3^x - 2^x \)

\( 3^0 - 2^0 = 0 \)

\( 3^1 - 2^1 = 3 - 2 = 1 = 1! \)

For x = 0, we get a 0 and then the rest of the sums will be non-zero. Curiously, for x = 1, the answer is 1!. This will become more obvious as we go higher.

\( 4^x - 2 \cdot 3^x + 2^x \)

\( 4^0 - 2 \cdot 3^0 + 2^0 = 1 - 2 + 1 = 0\)

\( 4^1 - 2 \cdot 3^1 + 2^1 = 4 - 6 + 2 = 0\)

\( 4^2 - 2 \cdot 3^2 + 2^2 = 16 - 18 + 4 = 2 = 2!\)

For the sum above, it took two values (x = 0 and x = 1) before we got a non-zero answer. And when x = 2, we get 2!.

\(5^x - 3 \cdot 4^x + 3 \cdot 3^x - 2^x \)

\(5^0 - 3 \cdot 4^0 + 3 \cdot 3^0 - 2^0 = 1 - 3 + 3 - 1 = 0 \)

\(5^1 - 3 \cdot 4^1 + 3 \cdot 3^1 - 2^1 = 5 - 12 + 9 - 2 = 14 - 14 = 0 \)

\(5^2 - 3 \cdot 4^2 + 3 \cdot 3^2 - 2^2 = 25 - 48 + 27 - 4 = 52 - 52 = 0 \)

\(5^3 - 3 \cdot 4^3 + 3 \cdot 3^3 - 2^3 = 125 - 192 + 81 - 8 = 206 - 200 = 6 = 3! \)

For the sum above, we get 3 zeros before we get 3!. This is a really interesting and amazing pattern.

\( 6^x - 4 \cdot 5^x + 6 \cdot 4^x - 4 \cdot 3^x + 2^x \)

\( 6^0 - 4 \cdot 5^0 + 6 \cdot 4^0 - 4 \cdot 3^0 + 2^0 = 1 - 4 + 6 - 4 + 1 = 0 \)

\( 6^1 - 4 \cdot 5^1 + 6 \cdot 4^1 - 4 \cdot 3^1 + 2^1 = 6 - 20 + 24 - 12 + 2 = 32 - 32 = 0 \)

\( 6^2 - 4 \cdot 5^2 + 6 \cdot 4^2 - 4 \cdot 3^2 + 2^2 = 36 - 100 + 96 - 36 + 4 = 136 - 136 = 0 \)

\( 6^3 - 4 \cdot 5^3 + 6 \cdot 4^3 - 4 \cdot 3^3 + 2^3 = 216 - 500 + 384 - 108 + 8 = 608 - 608 = 0 \)

\( 6^4 - 4 \cdot 5^4 + 6 \cdot 4^4 - 4 \cdot 3^4 + 2^4 = \) \( 1296 - 2500 + 1536 - 244 + 16 = \) \( 2848 - 2824 = 24 = 4! \)

For the sum above, we get 4 zeros before we get 4!.

\( 7^x - 5 \cdot 6^x + 10 \cdot 5^x - 10 \cdot 4^x + 5 \cdot 3^x - 2^x \)

\( 7^0 - 5 \cdot 6^0 + 10 \cdot 5^0 - 10 \cdot 4^0 + 5 \cdot 3^0 - 2^0 = \) \( 1 - 5 + 10 - 10 + 5 - 1 = 0 \)

\( 7^1 - 5 \cdot 6^1 + 10 \cdot 5^1 - 10 \cdot 4^1 + 5 \cdot 3^1 - 2^1 = \) \( 7 - 30 + 50 - 40 + 15 - 2 = 72 - 72 = 0 \)

\( 7^2 - 5 \cdot 6^2 + 10 \cdot 5^2 - 10 \cdot 4^2 + 5 \cdot 3^2 - 2^2 = \) \( 49 - 180 + 250 - 160 + 45 - 4 = 344 - 344 = 0 \)

\( 7^3 - 5 \cdot 6^3 + 10 \cdot 5^3 - 10 \cdot 4^3 + 5 \cdot 3^3 - 2^3 = \) \( 343 - 1080 + 1250 - 640 + 135 - 8 = 1728 - 1728 = 0 \)

\( 7^4 - 5 \cdot 6^4 + 10 \cdot 5^4 - 10 \cdot 4^4 + 5 \cdot 3^4 - 2^4 = \) \( 2401 - 6480 + 6250 - 2560 + 405 - 16 = 9056 - 9056 = 0 \)

\( 7^5 - 5 \cdot 6^5 + 10 \cdot 5^5 - 10 \cdot 4^5 + 5 \cdot 3^5 - 2^5 = \) \( 16807 - 38880 + 31250 - 10240 + 1215 - 32 = \) \( 49272 - 49152 = 120 = 5! \)

And the pattern is now obvious. Pascal’s Triangle has an amazing link with the coefficients for the power series sum formula. This formula is given here: Sum of the Power Series Formula.