Dear Uncle Colin,

I need to prove that $2\times 2^1 + 3\times 2^2 + 4\times 2^3 + \dots 2026\times 2^{2025}$ is a multiple of 2025 and I don’t know where to start!

- Finding A Calculation To Obviate Remainders

Hi, FACTOR, and thanks for your message! That looks fun.

Whenever I see something like that – a power series with lots of 2s in it, or similar – I like to replace the powers with an $x$, like this: $S = 2x + 3x^2 + 4x^3 + \dots 2026x^{2025}$. That immediately reminds me of the binomial expansion for $(1-x)^{-2}$, which is $1 + 2x + 3x^2 + \dots $ – with the problems that (a) that goes on forever and (b) it starts at 1. However, it’s a hint: I suspect that multiplying by $(1-x)$ twice might make this simpler.

If we multiply it by $(1-x)$, we get $2x + x^2 + x^3 + \dots - 2026x^{2026}$. That’s already a bit nicer!

Let’s multiply again, which gives us $2x - x^2 - 2027x^{2026} + 2026x^{2027}$ – all of the middle terms vanish! Meanwhile, (knowing that $x=2$, so $1-x = -1$), we haven’t actually changed the value of the expression at all by these multiplications.

Now the question is, how do we know this is a multiple of 2025? Well, the first two terms are 4 and -4, so we can ignore those and focus on the remaining $2026x^{2027} - 2027x^{2026}$. Again, noticing that $x=2$, that’s the same as $2026\times 2\times 2^{2026} - 2027\times ^{2026}$, or $(4052 - 2027)2^{2026}$. That first factor is 2025, so we’re done – the whole thing is equal to $2025\times 2^{2026}$.

Hope that helps!

- Uncle Colin