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    Sum of Consecutive Integers

    Let n be an integer greater than 1, show that 2n is the sum of two consecutive odd integers. Furthermore, show that 3n is the sum of three consecutive integers.

    Table Of Content

    1. There exists two consecutive odd integers whose sum is

    2. There exists three consecutive integers whose sum is

    These can all be easily proven via construction.

    1. There exists two consecutive odd integers whose sum is 2^n

    If there exists two odd consecutive odd integers satisfying the above property, then it must be the case that \exists m such that

    (1)   \begin{align*} (2m + 1) + (2m + 3) &= 2^n \\ 4m + 4 &= 2^n & n > 1\\ m + 1 &= 2^{n-2} \\ m &= 2^{n-2} - 1 &m\in \bf Z \end{align*}

    Since m = 2^{n-2} - 1 \in \bf Z, then 2m+1 = 2^{n-1} - 1 and 2m+3 = 2^{n-1}+1 are consecutive integers whose sum 2^{n-1} - 1 + 2^{n-1} + 1 = 2^n satisfies the above properties. \blacksquare

    2. There exists three consecutive integers whose sum is 3^n

    Suppose there exists three consecutive integers k,k+1,k+2 whose sum is 3k + 3 = 3^k, then because 3^k is necessarily odd, it must be the case that 3k + 3 must also be odd as well. Since the sum of an even number and an odd number is still odd, k must be even.

    (2)   \begin{align*} 3k + 3 &= 3^n \\ k + 1 &= 3^{n-1} & n > 1 \\ k &= 3^{n-1} - 1 & k \in \bf Z \\ \end{align*}

    Hence, the sum of 3^{n-1} - 1, 3^{n-1}, 3^{n-1} + 1 is 3^{n-1+1} = 3^n \blacksquare

    In fact, you can easily show that

        \[(4^{n-1}-3) + (4^{n-1}-1) + (4^{n-1}+1) + (4^{n-1}+3) = 4^n\]

    and that

        \[(5^{n-1}-2) + (5^{n-1}-1) + 5^{n-1}+ (5^{n-1}+1) + (5^{n-1}+2) = 5^n\]

    etc. Hence, using similar constructions, one can show that given an expression of the form k^n: if k is odd, then there exists k consecutive integers whose sum is k^n; on the other hand, if k is even, then there exists k consecutive odd integers whose sum is k^n.

    Posted By Lee

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