Prove that $$(2m-1)^m=\sum_{j=1}^m(2j-1)C^m_j(2m-1)^{m-j},$$ where $C^m_j$ denotes binomial coefficient. I tried induction but got nowhere. I guess some simple binomial coefficient identity will do the trick, but I cannot think of any. Thanks in advance

I believe that the following is true: $$\frac{d^n}{dx^n}f(x)g(x)=\sum_{i=0}^{\infty}\frac{n!}{i!(n-i)!}f^{(n-m)}(x)g^{(m)}(x)$$ The rational part of the summation is binomial expansion constants and $f^{(n)}(x)=\frac{d^n}{dx^n}f(x)$ I have tested it for some values of $n$ where $f$ and $g$ are either polynomials or exponential functions and it appears to hold true. The question is whether or not the above is true […]

How does one show that $\frac{\text{d}}{\text{d}x} x^n =nx^{n-1}$ without resorting to the Binomial Theorem? Edited : I’m interested in this approach as I’ve been tinkering with the proof for the Binomial Theorem using Taylor Series but then later realized that I’ve assumed $\frac{\text{d}}{\text{d}x} x^n = nx^{n-1}$ which I have proven earlier using the Binomial Theorem […]

This question already has an answer here: Evaluating even binomial coefficients 5 answers Alternating sum of binomial coefficients: given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$ 7 answers

Did i prove the Binomial Theorem correctly? I got a feeling I did, but need another set of eyes to look over my work. Not really much of a question, sorry.

Find the coefficient of: $[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$ I have figured out that I need to use this identity: $(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $ But I have no clue how to proceed with this, I have been stuck with this for hours please help.

I wanted to read a proof for the Binomial theorem, so I googled “proof of the binomial theorem“. My question is about the proof from the top link of that search. In the sixth line of the induction step, $$\sum_{k=0}^{n-1}{n-1 \choose k}x^{n-k}y^{k}$$ becomes $$\sum_{k=0}^{n}{n-1 \choose k}x^{n-k}y^{k}-{n-1 \choose n}x^{0}y^{n}$$ This is in order to get the […]

I’m trying to prove the following: For every $n \ge 5$: $$\sum_{k=0}^n{n\choose k}\left(-1\right)^k\left(n-k\right)^4 = 0$$ I’ve tried cancelling one $(n-k)$, and got this: $$n\sum_{k=0}^{n-1}{n-1\choose k}\left(-1\right)^k\left(n-k\right)^3 = 0$$ I’ve also tried expressing the first formula as such: $$\sum_{k=0}^na_kb_k$$ Where $a_k = {n \choose k}\left(-1\right)^k$ and $b_k = \left(n-k\right)^4 = \sum_{j=0}^4{4\choose j}n^j\left(-k\right)^{4-j}$ It’s easy to see that […]

Let (n; a,b,c) = n!/(a!*b!*c!) In other words, (n; a, b, c) is the trinomial coefficient. I am trying to find the triplets (a,b,c) which maximize this trinomial coefficient. I have determined, by simply plugging in numbers, that: When n mod 3 = 0, the trinomial coefficient is maximized when a = b = c […]

This question already has an answer here: A proof of $\sum_{i=1}^{n}(-1)^{i-1}\binom{n}{i}\frac{1}{i}=\frac{1}{1}+\frac{1}{2}+…+\frac{1}{n}$ [duplicate] 2 answers

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