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Use mathematical induction to prove that the sum of the entires of the $k$-th row of Pascal’s Triangle is $2^k$. Begin by proving that the row sum for any particular row is double that for the previous row.

I am having a hard time trying to figure out how to prove that the row sum for any particular row is double that for the previous row. I know how to show for row one, row two and so forth but once I get to row n I know that the sum has to be row(n-1)(2), but I have no idea how to prove that. I know that each row's sum can be written as $2^k$ where $k$ is the row number. I was wondering if anyone can give me a hint or start me off.

3 Answers 3

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Assume it for k. For k+1, We see that the $ k+1$ th row has two 1's and the rest other values are made by adding the numbers of the kth row two times except the ones on the corners of the kth row. So the sum of the numbers except the two ones of the kth row is $2^k-2$. So the sum of the k +1 th row is $ 1 + 1 + 2(2^k -2) + 1 +1 = 2^{k+1}$. Hence proved.

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To prove that each row sum is twice the previous row sum: given a particular row with values $(r_1,\dots, r_n)$, with $r_1=r_n=1$, then note that the next row is given by

$$(1,r_1+r_2,r_2+r_3,\dots, r_{n+1}+r_n,1) = (r_1,r_1+r_2,r_2+r_3\dots,r_{n+1}+r_n,r_n).$$

If we sum the elements of this row, we see that each $r_i$ appears exactly two times, so the row sum is $2 \sum_{i=1}^n r_i$; that is it is twice the previous row sum.

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There problem go from $n$ to $n+1$:

$$S_{n+1}={n+1\choose 0}+{n+1\choose 1}+...+{n+1\choose n+1}$$

Using Stiefel's rule:

$${n+1\choose k}={n\choose k}+{n\choose k-1}$$

and note that

$${n+1\choose 0}={n\choose 0}, \quad {n+1\choose n+1}={n\choose n}$$

so,

$$S_{n+1}={n\choose 0}+\left[{n\choose 0}+{n\choose 1}\right]+\left[{n\choose 1}+{n\choose 2}\right]+...+\left[{n\choose n-1}+{n\choose n}\right]+{n\choose n}=\\ =2\cdot \left[{n\choose 0}+{n\choose 1}+...+{n\choose n}\right]=2\cdot 2^n=2^{n+1}$$