Suppose the expression is given like this: $(1+x) ^{10} (1+x)^{20}$. How can I find out the coefficient of $x^m$ in the above expression, given that $0≤m≤20$.

Let $p\in \mathbb{R}$ and $n\in \mathbb{N}$ and $$\binom{p}{n}=\frac{p(p-1)(p-2)…(p-n+1)}{n!}$$ b) Prove $$n\binom{p}{n}=p\binom{p-1}{n-1}$$ Thanks for all the help with a! I definitely understand it now. Now part b is simply asking me to multiply those, correct? I did that and it just seems to easy. Also I never used the fact that $$\binom{p}{0}=1.$$ Was I supposed to […]

This question already has an answer here: Evaluate $\sum_{n=1}^\infty nx^{n-1}$ 2 answers

Find the term independent of $x$ and $y$ in $$\left(x+y+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}+2\right)^{20}$$ My attempt factorised it as$$\frac{(x+y+2)(xy+1)-1)}{xy}$$ but still the method is too lengthy

Expression I have somehow stumbled upon this expression (I believe I have proved it, but that is not important right now), which I have tried to simplify by writing it like something like this (I have been playing with powers of natural numbers and something similar to Pascal’s triangle) : n! = (t-1)^n , t^a=(a+1)^n […]

While playing around with random binomial coefficients , I observed that the following identity seems to hold for all positive integers $n$: $$ \sum_{k=0}^{2n} (-1)^k \binom{4n}{2k}\cdot\binom{2n}{k}^{-1}=\frac{1}{1-2n}.$$ However, I am unable to furnish a proof for it ( though this result is just a conjecture ). Any ideas/suggestions/solutions are welcome.

Questions: (1) Calculate $$\lim_{n \to \infty}\binom{2n}{n}$$ (2) Calculate $$\lim_{n \to \infty}\binom{2n}{n} 2^{-n}$$ without using L’Hôpital’s rule. Attempted answers: (1) Here I start by using the definition of a binomial: $$\lim_{n \to \infty}\binom{2n}{n} = \lim_{n \to \infty} \frac{(2n)!}{n!(2n-n)!} = \lim_{n \to \infty} \frac{(2n)!}{n!n!} = \lim_{n \to \infty} \frac{2n \cdot (2n-1) \cdot (2n-2) \cdot … \cdot (n […]

I’m undergraduate student of mathematics. I need to prove: $$\sum_{k=0}^{n} \binom{2n}{2k}= 2^{2n-1}$$ Can you please help me

This question already has an answer here: Dobble card game – mathematical background [duplicate] 1 answer

Let $$a=\frac{72!}{(36!)^2}-1$$ Find whether $a$ is odd. $a$ is even. $a$ is divisible by 71. $a$ is divisible by 73. Multiple answers can be correct. I was able to find whether $a$ is even or odd in the following way: There is a method of finding the power of a prime factor in the factorial […]

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