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In this video, James Grime shows that the number of $\mathbb{N}$ less than $10^{n+1}$ that have at least one $3$ among their digits is given by this recurrence relation:

$$T_{n+1} = 9T_{n} + 10^{n}\; where\; T_0=1$$

But later in the video, he says it can also be written as:

$$T_n = 10^{n+1} - 9^{n+1}$$

So, I set out to prove they are the same algebraically:

Solving the recurrence relation, I got:

$$T_n = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$

So, can somebody show me a proof to show that:

$$10^{n+1} - 9^{n+1} = 9^n + 9^{n-1}10 + 9^{n-2}10^2 + ... + 10^n$$

Truth-seek
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1 Answers1

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With $a^0=1$, $b^0=1$ for $a, b\ne 0$:

$(a-b)(a^{n}b^{0}+a^{n-1}b^{1}+a^{n-2}b^2+...+a^{1}b^{n-1}+a^{0}b^{n})$

$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-(a^{n}b^{1}+a^{n-1}b^{2}+...+a^{1}b^{n}+a^{0}b^{n+1})$

$=a^{n+1}b^{0}+a^{n}b^{1}+a^{n-1}b^2+...+a^{2}b^{n-1}+a^{1}b^{n}-a^{n}b^{1}-a^{n-1}b^{2}-...-a^{1}b^{n}-a^{0}b^{n+1}$

$=a^{n+1}b^{0}+a^{n}b^{1}-a^{n}b^{1}+a^{n-1}b^{2}-a^{n-1}b^{2}+...+a^{1}b^{n}-a^{1}b^{n}-a^{0}b^{n+1}$

$=a^{n+1}-b^{n+1}$

This will apply for your problem, because $a-b=10-9=1.$

user061703
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