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Prove by mathematical induction:

$\sum_{i=0}^{n}{n \choose i} = 2^n ; n \ge 0$

Step 1: n = 0

${0 \choose 0}=2^0$

Step 2:

for n = k

$\sum_{i=0}^{k}{k \choose i} = 2^k$

assumption: for n = k+1

$\sum_{i=0}^{k+1}{k+1 \choose i} = 2^{k+1}$

Step 3:

$\sum_{i=0}^{k+1}{k+1 \choose i} = \sum_{i=0}^{k}{k \choose i} + {k+1 \choose k+1} = 2^k+1$

Assumption:

$2^{k+1}$

Result of mathematical induction:

$2^k+1$

Is that correct or is somewhere the mistake ?

Martin Sleziak
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user288083
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    Your base case was good. But after the base case you are supposed to assume it holds true for $n = k$ and prove it holds true for $n = k + 1$. I think that's where you were going, but you wrote "assumption: for $n = k + 1$" which isn't an assumption you should be making – layman Nov 08 '15 at 22:20
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    This line is false : $\sum_{i=0}^{k+1}{k+1 \choose i} = \sum_{i=0}^{k}{k \choose i} + {k+1 \choose k+1}$. – C. Falcon Nov 08 '15 at 22:22
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    This is not written well; you should be able to read it as though it were just prose, and it should make grammatical sense. Try rewriting it as though you were explaining it to someone else. – Eric Tressler Nov 08 '15 at 22:22
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    In applying the inductive hypothesis, you assume that

    $$\sum_{k=0}^n {n \choose k} = 2^n$$

    and try to prove that

    $$\sum_{k=0}^{n+1} {n+1 \choose k} = 2^{n+1}.$$

    Usually, to make the inductive hypothesis helpful, you have to make the desired statement look more like the inductive hypothesis itself. One way your statements differ is that the ranges of summation are different. We can fix that: ${n+1\choose n+1}=1$, so

    $$\sum_{k=0}^{n+1} {n+1 \choose k} = 1+\sum_{k=0}^n {n+1 \choose k}.$$

    But now you need to do something to make ${n+1 \choose k}$ "look like" ${n \choose k}$ when $k \leq n$.

     – Ian Nov 08 '15 at 22:24
  • I am already know where is mistakes. But I dont know how to proceed now ? How to modify this statement: $\sum_{k=0}^{n+1} {n+1 \choose k} = 1+\sum_{k=0}^n {n+1 \choose k}....$ Can anyone help mi please? – user288083 Nov 08 '15 at 22:46
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    I found the topic with same assignment and solving link – user288083 Nov 08 '15 at 23:04
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    @user288083 We have the "Pascal's triangle" identity ${n+1 \choose k}={n \choose k}+{n \choose k-1}$ whenever $k$ is between $1$ and $n$ inclusive. (Try to prove this.) Consequently $\sum_{k=0}^{n+1} {n+1 \choose k} = {n+1 \choose 0} + {n+1 \choose n+1} + \sum_{k=1}^n {n \choose k} + \sum_{k=1}^n {n \choose k-1}$. Can you evaluate these two sums using your inductive hypothesis? – Ian Nov 08 '15 at 23:05
  • Use the fact that ${n\choose k} +{n\choose k+1} = {n+1\choose k+1}$. For example, If you assume ${1\choose 0} +{1\choose 1} =2^1$, then ${2\choose 0} +{2\choose 1}+{2\choose 2} = ({1\choose 0}+0) +({1\choose 0}+{1\choose 1}) +({1\choose 1} + 0) = 2({1\choose 0} + {1\choose 1})=2\cdot2^1=2^2$ – EA304GT Nov 08 '15 at 23:08
  • If the main purpose of your post is to check your proof, then you should tag it ([tag:proof-verification]) or ([tag:solution-verification]). If you are simply interested in inductive proof of this fact, then there are many other posts on this site with such proof. – Martin Sleziak Nov 09 '15 at 16:24

1 Answers1

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Your base case is essentially correct, although I would prefer to see the justification that both sides are equal to $1$. In step 2, your assumption should be $$\sum_{i = 0}^{k} \binom{k}{i} = 2^k$$ since you must show that $P(k + 1)$ holds whenever $P(k)$ holds. In step 3, the statement $$\sum_{i = 0}^{k + 1} \binom{k + 1}{i} = \sum_{i = 0}^{k} \binom{k}{i} + \binom{k + 1}{k + 1}$$ is false. Furthermore, if $k$ is a positive integer, $2^{k + 1} = 2 \cdot 2^k = 2^k + 2^k > 2^k + 1$.

An induction proof is provided below:

Proof. Let $P(n)$ be the statement that $$\sum_{k = 0}^{n} \binom{n}{k} = 2^n$$

Let $n = 0$. Then $$\sum_{k = 0}^{0} \binom{0}{k} = \binom{0}{0} = 1 = 2^0$$ Hence, $P(0)$ holds.

Since $P(0)$ holds, we may assume there exists a nonnegative integer $m$ such that $P(m)$ holds. Then $$\sum_{k = 0}^{m} \binom{n}{k} = 2^m$$ Let $n = m + 1$. We must show that $P(m) \Rightarrow P(m + 1)$.
\begin{align*} \sum_{k = 0}^{m + 1} \binom{m + 1}{k} & = \binom{m + 1}{0} + \sum_{k = 1}^{m} \binom{m + 1}{k} + \binom{m + 1}{m + 1}\\ & = 1 + \sum_{k = 1}^{m} \binom{m + 1}{k} + 1\\ & = 1 + \sum_{k = 1}^{m} \left[\binom{m}{k} + \binom{m}{k - 1}\right] + 1 & \text{by Pascal's Identity}\\ & = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{k = 1}^{m} \binom{m}{k - 1} + 1\\ & = 1 + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + 1 & \text{where $j = k - 1$}\\ & = \binom{m}{0} + \sum_{k = 1}^{m} \binom{m}{k} + \sum_{j = 0}^{m - 1} \binom{m}{j} + \binom{m}{m}\\ & = \sum_{k = 0}^{m} \binom{m}{k} + \sum_{j = 0}^{m} \binom{m}{j}\\ & = 2^m + 2^m & \text{induction hypothesis}\\ & = 2 \cdot 2^{m}\\ & = 2^{m + 1} \end{align*} Since $P(0)$ holds and $P(m) \Rightarrow P(m + 1)$ for each nonnegative integer $m$, $P(n)$ holds for all nonnegative integers.$\blacksquare$

Note. The binomial coefficient $$\binom{n}{k}$$ represents the number of subsets of size $k$ that can be selected from a set of size $n$. Thus, the summation $$\sum_{k = 0}^{n} \binom{n}{k}$$ counts the number of subsets of a set with $n$ elements. There are $2^n$ such subsets since a subset is determined by the decision of whether or not to include each of the $n$ elements.

N. F. Taussig
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