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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)$ $\binom{n}{0}-\binom{n}{1}+\binom{n}{3}-\cdots+(-1)^n\binom{n}{n}=0, \forall n\in \mathbb N$

$ iii)$ $\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}+\cdots+n\binom{n}{n}=n(2^{n-1}), \forall n\in \mathbb N$

$ iv)$ $1\cdot 2\binom{n}{2}+2\cdot 3\binom{n}{3}+\cdots+(n-1)\cdot n\binom{n}{n}=n(n-1)2^{n-2}, \forall n\in \mathbb N$

I really don't know how I can start with this.

Martin Sleziak
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Victor F Salazar G
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    You can start by checking the base case of the induction. Then you should try to do the induction step. How can you simplify the corresponding terms for $n+1$? – Listing Jan 08 '14 at 17:33
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    What exactly is newton's formula here ? I looked at wikipedia, it seems to deal with sums of binomials. Maybe you do not need induction to prove some of the given formulas. – Peter Jan 08 '14 at 17:39
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    If it is simply the classic binomial theorem, you could solve i) by simple setting a=b=1. – Peter Jan 08 '14 at 17:43
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    And ii) could be solved by setting a=1 , b=-1. – Peter Jan 08 '14 at 17:44
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    to prove all these formulas no need induction – Leox Jan 08 '14 at 18:09
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    An easier argument for this is a combinatorial proof.. – ireallydonknow Jan 08 '14 at 18:37
  • Please, post only one question in one post. Posting several questions in the same post is discouraged and such questions may be put on hold, see meta. – Martin Sleziak Jan 09 '14 at 08:35
  • Your third question has been asked also here: http://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r-0n-r-n-choose-r-n2n-1 (Although that question does not ask specifically for the inductive proof.) – Martin Sleziak Jan 09 '14 at 08:39
  • The first sum is solved here: http://math.stackexchange.com/questions/177405/prove-by-induction-2n-cn-0-cn-1-cdots-cn-n See also this question: http://math.stackexchange.com/questions/27539/proving-a-special-case-of-the-binomial-theorem – Martin Sleziak Jan 09 '14 at 08:43
  • This is not a question, this is about four questions! How did you manage to post them all as one? – Gigili Jan 09 '14 at 09:32

2 Answers2

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This feels like your homework and I'm not comfortable simply solving you the problems. I'm not sure if you already know that, but I will simply explain and provide examples of the needed subjects. You can report any specific problems you are having:

Proof by induction

You can use that to proof statements about natural numbers. Identify the statement $P(n)$ you wish to prove and do the following:

1) Prove the base case, ie, the smallest natural number for which the statement is true.

2) Suppose P(n) is true and prove that it implies in P(n+1) being true.

Example: $1 + 2 + 3 + ... + n = \frac{(n)(n+1)}{2}$

It is clearly true for n = 0. Supposing it is true for n, we have:

$1 + 2 + 3 + ... + n + (n+1) = \frac{(n)(n+1)}{2} + (n+1) = \frac{(n)(n+1) + 2(n+1)}{2} = \frac{(n+1)(n+2)}{2}$

Which is the statemente for $n+1$. This ends the proof.

Newton's binomial formula

I won't prove it. I will simply show and use one or two times.

$(a + b)^n = \sum_{i=0}^{n} a^ib^{n-i}\binom{n}{i}$

For example:

$1.04^2 = (1 + 0.04)^2 = 1^0\times0.04^2 \binom{2}{0} + 1^1\times0.04^1 \binom{2}{1} + 1^2\times0.04^0 \binom{2}{2} = 1.0816$

Of course this is a very complicated way of doing something simple. But the formula should help in several of your exercices. Try to apply for the correct values of $a$ and $b$.

lsoranco
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Here is an alternative approach (without Newton's Binomial):

Let's take an $n$-bit zero binary string. There are $\binom{n}{k}$ ways to choose $k$ bits and turning them on. Since the total number of ways of choosing how to turn them on is $2^n$, but at the same time is the number of ways of turning $0,1,2,...n$ bits on, we are done.

chubakueno
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