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I'm looking for a proof verification, precisely at the end of the proof. We proceed by induction on $n$.

Pf. Base case $n = 1$,$$\sum_{j=0}^1 {1 \choose j} = {1 \choose 0} + {1 \choose 1} = 1 + 1 = 2^1$$ Notice also for $n=2$, $$\sum_{j=0}^2 {2 \choose j} = {2 \choose 0} + {2 \choose 1} + {2 \choose 2} = 1 + 2 + 1 = 2^2$$ Now, suppose that for $k\in\mathbb{N}$ it is true that $$\sum^k_{j=0} {k \choose j} = 2^k$$ Then, by Pascal's identity, $$\sum^k_{j=0} {k \choose j} + \sum^k_{j=0} {k \choose j-1} = \sum^{k+1}_{j=0} {k+1 \choose j} = 2^{k+1}$$ Since $k$ is any natural number, it should then also work for $k+1$.

My reservations are that I'm using the assumption incorrectly or misusing the assumption by not starting with the assumption. Any clarity is appreciated.

Edit: It seems that changing the bound on the second sum in this equation

Then, by Pascal's identity, $$\sum^k_{j=0} {k \choose j} + \sum^k_{j=0} {k \choose j-1} = \sum^{k+1}_{j=0} {k+1 \choose j} = 2^{k+1}$$

to $k+1$ yielded the quickest fix to the original proof, with the convention that, as Brian M. Scott pointed out: ${n \choose k}$ is 0 if $k$ is a negative integer.

dan
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  • Looks like a proper induction proof to me – Dhanvi Sreenivasan Jun 12 '20 at 03:39
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    Your mistake is that, if $j=0$ then first term will be ${k\choose 0}+{k\choose {-1}}$; where factorial of $-1$ is not defined. – SarGe Jun 12 '20 at 03:44
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    @Doubtnut: There is no problem: $\binom{n}k$ is defined to be $0$ if $k$ is a negative integer and $$\frac{n^{\underline{k}}}{k!}=\frac{n(n-1)(n-2)\ldots(n-k+1)}{k!}$$ if $k$ is a non-negative integer. – Brian M. Scott Jun 12 '20 at 04:00
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    @BrianM.Scott It is the usual convention to have $\binom nk=0$ whenever $k$ is out of range, but the (minor) problem with the proposer's solution is that his first two sums end at $j=k$. The second needs to go up to $j=k+1$. – Angina Seng Jun 12 '20 at 04:22
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    @AnginaSeng: Yes, that, unlike the negative lower number, is a genuine (though indeed minor) problem. – Brian M. Scott Jun 12 '20 at 04:31
  • @AnginaSeng Changing the upper bound of the second sum cleared up the proof. +1 – dan Jun 12 '20 at 05:23

2 Answers2

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Your argument is incorrect.

First you state:

$$\sum^k_{j=0} {k \choose j} = 2^k$$

and then claim:

$$\sum^k_{j=0} {k \choose j} + \sum^k_{j=0} {k \choose j-1} = \sum^{k+1}_{j=0} {k+1 \choose j} = 2^{k+1}$$

However:

$$\sum^k_{j=0} {k \choose j-1}=\sum^k_{j=0} {k \choose j}-{k\choose k}=2^k-1$$

so your LHS = $2^{k+1}-1$.

JMP
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To salvage your proof, use Pascal's identity in $$ \sum^{k+1}_{j=0} {k \choose j} + \sum^{k+1}_{j=0} {k \choose j-1} = \sum^{k+1}_{j=0} {k+1 \choose j}. $$ Both summands on the LHS are then equal to $2^k$, by your induction hypothesis, with the aid of the convention ${k\choose j}=0$ when $j<0$ or $j>k$.

grand_chat
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  • This is the most helpful answer. It seems that changing the upper bound of the second sum to $k+1$ in the original post is a variation of what you have posted and so this is the closest. Thank you. – dan Jun 12 '20 at 05:37