Sedrakyan's Power Sums Triangle

A self-constructive Pascal's type triangle — Hayk Sedrakyan

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Sums of the form 1k+2k+...+nk1^{k} + 2^{k} +... + n^{k}, have been studied for centuries, where nn and kk are positive integers. We introduce a self-constructive, Pascal-type triangle. This operates similarly to how Pascal's triangle handles expansions of the form (a+b)n(a+b)^{n}, offering an approach despite the existence of Newton's binomial theorem.

Self-Construction Principle (see the figure below: Sedrakyan's Power Sums Triangle).

The construction of the triangle is governed by the following rules:

  • Initial Condition: The first row starts with 12\dfrac{1}{2} and 12\dfrac{1}{2}, representing the formula for the sum of the first nn integers:
    1+2+⋯+n=n(n+1)2=12⋅n+12⋅n21 + 2 + \dots + n = \frac{n(n+1)}{2} = \dfrac{\mathbf{1}}{\mathbf{2}} \cdot n + \dfrac{\mathbf{1}}{\mathbf{2}} \cdot n^{2}
  • Rightmost Leading Diagonal: The coefficients along this diagonal are equal 1k+1\dfrac{1}{k+1}.
  • First Rightmost Dashed Diagonal: The coefficients along this diagonal remain constant at 12\dfrac{1}{2}.
  • Row Initialization: Subsequent rows start with BkB_{k}, where BkB_{k} are the Bernoulli numbers.
  • Recursive Step: New row terms are built recursively by multiplying the previous diagonal term by kp\dfrac{k}{p}, where kk is the row index and pp is the power of nn.

Bk=0B_{k} = 0 for odd k>1k > 1, so odd rows begin with 0 while even rows start with B2=16, B4=−130, B6=142,...B_{2} = \dfrac{1}{6},\ B_{4} = -\dfrac{1}{30},\ B_{6} = \dfrac{1}{42},...

Sedrakyan's Power Sums Triangle

Sedrakyan's power sums triangle

Application. Example (k=4k = 4)

Calculate the sum 14+24+...+n41^{4} + 2^{4}+...+n^{4}.

Solution. We have

14+24+...+n4=15⋅n5+12⋅n4+B2⋅32⋅43⋅n3+0⋅n2+B4⋅n.1^{4} + 2^{4}+...+n^{4} = \dfrac{1}{5} \cdot n^{5} + \dfrac{1}{2} \cdot n^{4} + B_{2} \cdot \dfrac{3}{2} \cdot \dfrac{4}{3} \cdot n^{3} + 0 \cdot n^{2} + B_{4} \cdot n.

Thus

14+24+...+n4=n55+n42+n33−n30=n(n+1)(2n+1)(3n2+3n−1)30.1^{4} + 2^{4}+...+n^{4} = \dfrac{n^{5}}{5} + \dfrac{n^{4}}{2} + \dfrac{n^{3}}{3} - \dfrac{n}{30} = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}.

References

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  2. Coolidge, J. L., A Historically Interesting Formula for the Area of a Quadrilateral. Amer. Math. Monthly 46, 345–347, 1939.
  3. Acerbi, F., Vitrac, B. Heron of Alexandria: Metrica, Heron's formula. Serra, 2014.
  4. Heath, T. L. A history of Greek mathematics (Vol. 2). Clarendon Press, 1921.
  5. H. Sedrakyan, N. Sedrakyan, Mathematics Competitions Must-Know Geometry: Practice and theory inspired by the style of AMC and AIME geometry (unofficial guide), 2023.
  6. H. Sedrakyan, N. Sedrakyan, Algebraic inequalities, Springer International Publishing, 2018.

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