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Proof of Arithmetic Summation Formula

On an intuitive level, the formula for the sum of a finite arithmetic series says that the sum of the entire series is the average of the first and last values, times the number of values being added.

    (n/2)(a-sub-1 + a-sub-n) = n [ (a-sub-1 + a-sub-n) / 2 ]

This makes sense, especially if you think of a summation visually as being the sum of the areas of the pictured bars. Copyright Elizabeth Stapel 2006-2011 All Rights Reserved

    n = 1, 2, 3,... and a_i = a_1, a_2, a_3,...

Since the bars grow by a fixed amount at each step, you can, in effect, "average" the bars to get the total area:

    animation: short bars grow, tall bars shrink, meeting in the middle at the average value

To prove this equality properly, however, requires a bit more work. We proceed by induction:

    Proof: Let n = 2. Then we have:

      a-sub-1 + a-sub-2 = (2/2)(a-sub-1 + a-sub-2)

    For n = k, assume the following:

      a_1 + a_2 + ... + a_k = (k/2)(a_1 + a_k)

    Let n = k + 1. Then we have:

      a_1 + a_2 + ... + a_k + a_(k+1) = (ka_1 + ka_k + 2a_(k+1)) / 2

    By nature of arithmetic sequences, we have:

      ak = ak+1  d
      +1 = a1 + kd

    Then, substituting the above into the n = k + 1 expression, we have:

      = (ka_1 + ka_(k+1) + a_1 + a_(k+1)) / 2 = [ (k + 1) / 2 ] [a_1 + a_(k+1)]

    Therefore the result holds for n = k + 1, and the formula is proved for all n > 2.

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Cite this article as:

Stapel, Elizabeth. "Proof of Arithmetic Summation Formula." Purplemath. Available from 6.htm. Accessed


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