If $(a+b)^n=\sum_{k=0}^{n}{n\choose k}a^{n-k}b^kc_k$, then $c_k=1$?

Question

Be advised this is a real soft question:

If $$(a+b)^n=\sum_{k=0}^{n}{n\choose k}a^{n-k}b^kc_k$$ Assuming $abc \neq 0$ must we have the following condition? $$c_k=1$$ for all $0 \leq k \leq n$

How do you prove it?

Answer 1

I tried to generate a function that equates to $0 $ for all $0 \leq k \leq n$ as followed $$f(n)=(a+b)^n-\sum_{k=0}^{n}{n\choose k}a^{n-k}b^kc_k=0$$ It turns out that $c_k$ does not have to be necessarily equal to 1 for that to happen. Funny, this only holds true for $n=0$. One can quickly verify for $n=0$, we obtain: $$f(0)=(a+b)^0-\sum_{k=0}^{0}{0\choose 0}a^{0-0}b^0c_0=0$$ $$f(0)=1-1c_0=0$$ which implies $$c_0=1$$ However, for all $n>0$, we have a different story. For instance, for $n=1$ $$f(1)=(a+b)-(ac_0+bc_1)=0$$ Where clearly we have infinitely many solutions for the following cases:$$c_0 = 1-\lambda b$$ $$c_1 =1+ \lambda a $$ where $\lambda $ is any integer.

Answer 2

Clarification: Based on the OP's response to my question in the comments above, I am assuming that the equation has to hold for all $n$. (If the statement is only true for a fixed $n$, then it is clear that the functional $c \mapsto \sum_{k=0}^{n}{n\choose k}a^{n-k}b^kc_k$ is linear and has a non trivial kernel for $n>0$, in which case the $c_k$ cannot be unique.)

Since $a\neq 0$ we can divide across by $a^n$, then letting $x={b \over a}$ we can write the condition as $\sum_{k=0}^n \binom{n}{k} x^k (c_k -1) = 0$.

Setting $n=0$ gives $c_0 = 1$. Suppose $c_k = 1$ for $k=0,...,n$, then $\sum_{k=0}^{n+1} \binom{n+1}{k} x^k (c_k -1) = \binom{n+1}{n+1} x^{n+1} (c_{n+1} -1) = 0$, from which we get $c_{n+1} = 1$.