3
votes
2answers
82 views

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 ...
2
votes
0answers
42 views

${n \choose r}=\frac {n!}{r!(n-r)!}$ 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 ...
1
vote
1answer
32 views

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 ...
1
vote
1answer
87 views

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 ...
3
votes
3answers
114 views

Binomial theorem $(a+b)^n=\sum \limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$ [duplicate]

I'm trying to understand the proof by induction of: $$ (a+b)^n = \sum \limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k} $$ I'm at the point of deriving the inductive step and am getting next: $$ (a+b)^{n+1} = ...
1
vote
2answers
54 views

Proving a summation involving binomial coefficients.

I need to prove the following inductively: (http://upload.wikimedia.org/math/9/e/5/9e57871ba17c1ad48e01beb7e1bb3bb9.png) $$\sum_{i=1}^{n} i{n \choose i} = n2^{n-1}$$ And for the life of me I can't ...
1
vote
2answers
200 views

Show $\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$

My question is: show $$\sum_{k=0}^{m} \binom{n-k}{m-k}=\binom{n+1}{m}$$ $$n\geq m\geq 1$$ I tried to do this via induction and failed. there has to be another way of doing this. We could either ...
1
vote
2answers
62 views

How to prove $\sum_{k\leq n}^{n} \binom{n}{k}= 2^n$ by induction [duplicate]

$\sum_{k\leq n}^{n} \binom{n}{k}= 2^n , n, k \in \mathbb{N}$ Im trying with mathematical induction but im stuck. My inductive step: $H) \sum_{k=0}^{h} \binom{h}{k}= 2^h$ $T) \sum_{k=0}^{h+1} ...
2
votes
2answers
175 views

Use induction and Newton's binomial formula to show that $\binom{n}{0}+\binom{n}{1}+\cdot+\binom{n}{n}=2^n, \forall n\in \mathbb N$ [duplicate]

Use induction and Newton's binomial formula to show that: $ i)$ $ \binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n, \forall n\in \mathbb N$ $ ii)$ ...
0
votes
1answer
97 views

Binomial coefficients identity (sum of the powers of the natural numbers)

I've found exercise with binomial coefficients in Kostrikin's book. Proof that $\sum_{i=1}^n{{r+1}\choose{i}}\left(1^i+2^i+\dots+n^i\right)=(n+1)^{r+1}-(n+1)$ I was trying to check that for ...
3
votes
3answers
185 views

Given n $\in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$

I tried to solve it using induction, but that got me no were, in the middle of the equation stat appearing ks that I don't see how to get out of the equation. I think the easiest way to prove it, it's ...
0
votes
1answer
65 views

Inequality with a sum and factorial

For a homework assignment we have the following question that I'm stuck on. Let $ 0 \leq y \leq 1 $ be given. $\forall m \in \mathbb{N}$, define $ \displaystyle S_m(y)=\sum_{k=0}^m \binom{m}{k}y^k$. ...
0
votes
1answer
182 views

Induction proof involving prime factor and binomial coefficients

The question in my book is: Use induction on $r$ to prove that if $p$ is a prime integer, then $p$ is a factor $\binom{p}{r}$ for $r=1, 2, . . . , p-1$. I'm not really sure how to go about ...
-1
votes
1answer
83 views

Revisited: Binomial Theorem: An Inductive Proof

I'm asked to use the fact that $\begin{pmatrix}n\\r\end{pmatrix}+\begin{pmatrix}n\\r-1\end{pmatrix}=\begin{pmatrix}n+1\\r\end{pmatrix}$ to show, by induction, that ...
4
votes
6answers
97 views

Is $\sum_{i=1}^{n-1}i=\binom{n}{2}$?

How can I show that $$ \sum_{i=1}^{n-1}i=\binom{n}{2}? $$ This is what I have tried, but I do not know if it is correct: Proof. Let $n=2$. Then, $$ \begin{align} \sum_{i=1}^{1}i&=1\text{, ...
2
votes
3answers
299 views

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 $m$?
3
votes
3answers
272 views

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 ...
6
votes
2answers
2k views

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 ...
2
votes
5answers
175 views

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 $n=0$: ...
2
votes
2answers
145 views

Prove by Induction

For $n\in \mathbb{N}$ and $z\in \mathbb{C}$: $\sin{(nz)}=\sum _{ k=0 }^{ n }{ \binom{n}{k} }\frac{1}{2i}(i^k-(-i)^k)(\cos{z})^{n-k}(\sin{z})^k $ $\cos{(nz)}=\sum _{ k=0 }^{ n }{ ...
17
votes
6answers
2k views

Induction: $\sum_{k=0}^n \binom nk k^2 = n(1+n)2^{n-2}$

I found crazy (for me at least) induction example, in fact it just would be nice to prove. (Even have problems with starting) Any hints are highly valued: ...
2
votes
2answers
95 views

Prove the following relation:

I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$ I got this far before I got stuck: $\begin{eqnarray*} ...
0
votes
5answers
147 views

Prove by induction: $2^n = C(n,0) + C(n,1) + \cdots + C(n,n)$ [duplicate]

This is a question I came across in an old midterm and I'm not sure how to do it. Any help is appreciated. $$2^n = C(n,0) + C(n,1) + \cdots + C(n,n).$$ Prove this statement is true for all $n ...
5
votes
2answers
456 views

Odd Binomial Coefficients?

By Newton's Formula: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k $$ Proof that every $\dbinom{n}{k}$ is odd if and only if $n=2^r-1$. I have already shown that if $n$ is of the form $2^r-1$, ...
3
votes
2answers
263 views

Lower Bound of Central Binomial Coefficients

I would like to prove by induction the following inequality: $\frac{4^n}{n+1} < \binom{2n}{n}$, for all natural numbers n > 1. Any hints?
4
votes
2answers
223 views

Proof of $(2n)!/(n!)^2\le2^{2n}$ by mathematical induction?

How do I approach this problem using mathematical induction? $$\frac{(2n)!}{(n!)^2} \leq 2^{2n}$$
0
votes
1answer
120 views

Algebra on $\binom{k+1}{i} = \binom{k+1}{0} + \binom{k+1}{1} + \cdots + \binom{k+1}{k} + \binom{k+1}{k+1}$ [duplicate]

Possible Duplicate: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$ I am trying to prove $\sum \limits_{i=0}^n \binom{n}{i} = 2^n$ by induction. I've been all over the net ...
1
vote
2answers
662 views

Induction proof concerning a sum of binomial coefficients: $\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$

I'm looking for a proof of this identity but where j=m not j=0 http://www.proofwiki.org/wiki/Sum_of_Binomial_Coefficients_over_Upper_Index $$\sum_{j=m}^n\binom{j}{m}=\binom{n+1}{m+1}$$
6
votes
3answers
1k views

Inductive proof for the Binomial Theorem for rising factorials

I want to proove the following equality containing rising factorials $$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$ For $n=1$ this equality is ...