# Tagged Questions

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### Prove by induction that $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ is decreasing

I want to prove that the following sequence is monotonously decreasing: $A_k = \sum\limits_{n=2k}^{3k}\binom{3k}{n}\cdot{(\frac{60}{100})}^n\cdot{(\frac{40}{100})}^{3k-n}$ I think it should be ...
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### ${n \choose r}=\frac {n!}{r!(n-r)!}$ without using the permutation approach.

I had an idea that would be to first prove Pascal's Rule, $${n \choose r} = {n-1 \choose r-1} + {n-1 \choose r}$$ which can be proved combinatorically whether one particular element(among the $n$) is ...
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### Binomial Coefficient Recusions

Let m and j be non-negative integers. Define $S^{0}_{m} = 1$ and: $S^{j}_{m} = \displaystyle\sum\limits_{i=1}^{m} S_{i}^{j-1}$ Show via induction: $S_{m}^{j} = {m+j-1 \choose j}$ I can ...
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### Can't find an identy for proving that $\sum_{k=0}^{i+1} \binom {i+1} k=2^{i+1}$ [duplicate]

$$\sum_{k=0}^{i+1} \binom {i+1} k$$ I can't find an identity for this summation :( To clarify I'm trying to prove using induction that this sum is equal to $2^{i+1}$, I have my basis and ...
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### combinatorial argument and by induction proof

Let n be a fixed natural number. Show that: $$\sum_{r=0}^m \binom {n+r-1}r = \binom {n+m}{m}$$ (A): using a combinatorial argument and (B): by induction on $m$?
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### Induction proof of $\sum_{j=0}^n{(-1)^j{n \choose j}\prod_{k=m+1}^{m+n-1}{(j+k)}}=0$

Does anynone have some hints to prove the following equation by induction for all $n\geq 1$ and $m\in\mathbb{Z}$ $$\sum_{j=0}^n{(-1)^j{n \choose j}\prod_{k=m+1}^{m+n-1}{(j+k)}}=0$$ use for ...
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### Inductive proof that ${2n\choose n}=\sum{n\choose i}^2.$

I would like to prove inductively that $${2n\choose n}=\sum_{i=0}^n{n\choose i}^2.$$ I know a couple of non-inductive proofs, but I can't do it this way. The inductive step eludes me. I tried naively ...
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### Induction on binomial Identity

I am having trouble proving the following identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ Here is what I have so far: Proof: Base: Let $n=0$: ...