Sums of Integers

John W. Dawson
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John W. Dawson: Penn State York

Chapter Chapter 3 in Why Prove it Again?, 2015, pp 13-18 from Springer

Abstract: Abstract As a first, very simple case study of alternative proofs, consider the following four proofs of the identity 1 + 3 + … + ( 2 n − 1 ) = n 2 $$1 + 3 +\ldots +(2n - 1) = n^{2}$$ (a result known since antiquity, and one that is readily conjectured from specific instances): 1. Proof via gnomons (Figure 3.1) 2. Proof by induction:The result is true when n = 1, since 1 = 12. Assuming that it is true when n = k, suppose n = k + 1 $$n = k + 1$$ . Then 1 + 3 + … + ( 2 k − 1 ) + ( 2 ( k + 1 ) − 1 ) = 1 + 3 + … + ( 2 k + 1 ) = k 2 + ( 2 k + 1 ) = ( k + 1 ) 2 $$1+3+\ldots +(2k-1)+(2(k+1)-1) = 1+3+\ldots +(2k+1) = k^{2}+(2k+1) = (k+1)^{2}$$ . By induction, the result therefore holds for all n. 3. Proof via Gauss’s method (recalling the tale of his schoolboy summation of the first hundred integers1): 1 + 3 + 5 + … + ( 2 n − 1 ) = S ( 2 n − 1 ) + … + 5 + 3 + 1 = S $$\displaystyle\begin{array}{rcl} 1 + 3 + 5 +\ldots +(2n - 1)& =& S {}\\ (2n - 1) +\ldots +5 + 3 + 1& =& S {}\\ \end{array}$$ whence, adding the equations together column by column, 2 n + 2 n + … + 2 n = 2 S $$2n + 2n +\ldots +2n = 2S$$ , where the left member of the equation has n equal summands. Thus 2n ⋅ n = 2S, or S = n 2. 4. The stairstep proof (Figure 3.2): Which of these proofs are to be regarded as essentially different, and on what grounds?

Keywords: Geometric Representation; Unit Cube; Mathematical Induction; Rectangular Parallelepiped; Inductive Proof (search for similar items in EconPapers)
Date: 2015
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DOI: 10.1007/978-3-319-17368-9_3

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