Thread: Binomial coefficients

Ever thought of that the binomial koefficients equal 11ⁿ?

Eh what?

The sum of binomial coefficients is equal to 2ⁿ. For example, ⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵.

Proof: Take x = 1 in the binomial expansion of (1+x)ⁿ.

Yeah, I thought that's perhaps what he meant.

I think he meant that the digits of 11^n are binomial coefficients.
1
11
121
1331
14641
161051 <- It breaks here though

101^n lasts a while longer. In general, any 10^k+1 works for a while and the reason is pretty simple. (a + b)^n = sum over i of (n choose i)*(a^i)*(b^(n-i)), so for a=10, b=1 you're just left with powers of 10 times the binomial coefficients.

You are right about what I meant Magi Master but the only reason that the binomial coefficients aren't palindroms all the time even if they equal 11ⁿ is that ten symbols, as we use, aren't enough.
There will be overflow.
If you want to you can write a computer program with for example 15 symbols and there you'll see that 11⁵ is a palindrom.

Yes, in a higher base, 11ⁿ will stay a palindrome for longer, but never indefinitely (101ⁿ is about the same as 11ⁿ in base-100). I suppose you could define something like a base-infinity (based on tuples) that would stay a palindrome but I don't think there's much point in that (correct me if I'm wrong).

The only point is that palindromes are beautiful.