Numbers whose Cube equals Sum of Sequence of that many Squares/Examples/47

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Example of Numbers whose Cube equals Sum of Sequence of that many Squares

$\ds 47^3 = \sum_{k \mathop = 1}^{47} \paren {21 + k}^2$


Proof

From Numbers whose Cube equals Sum of Sequence of that many Squares:

$\ds m^3 = \sum_{k \mathop = 1}^m \paren {n + k}^2$

for $m = 47$ and for $n$ given by:

$n = \dfrac {m + 1} 2 \pm \dfrac 1 6 \sqrt {33 m^2 + 3}$


So in this instance:

\(\ds n\) \(=\) \(\ds -\dfrac {47 + 1} 2 \pm \frac 1 6 \sqrt {33 \times 47^2 + 3}\)
\(\ds \) \(=\) \(\ds -24 \pm \frac 1 6 \sqrt {33 \times 2209 + 3}\)
\(\ds \) \(=\) \(\ds -24 \pm \frac 1 6 \sqrt {72 \, 900}\)
\(\ds \) \(=\) \(\ds -24 \pm \frac {270} 6\)
\(\ds \) \(=\) \(\ds -24 \pm 45\)


This gives:

$n = 21, n = -69$

which leads to the two solutions:

\(\ds 47^3\) \(=\) \(\ds 22^2 + 23^2 + \cdots + 68^2\)
\(\ds \) \(=\) \(\ds \paren {-68}^2 + \paren {-67}^2 + \cdots + \paren {-22}^2\)


Hence we have:

\(\ds \sum_{k \mathop = 1}^{47} \paren {21 + k}^2\) \(=\) \(\ds \sum_{k \mathop = 22}^{68} k^2\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^{68} k^2 - \sum_{k \mathop = 1}^{21} k^2\)
\(\ds \) \(=\) \(\ds \frac {68 \paren {68 + 1} \paren {2 \times 68 + 1} } 6 - \frac {21 \paren {21 + 1} \paren {2 \times 21 + 1} } 6\) Sum of Sequence of Squares
\(\ds \) \(=\) \(\ds \frac {68 \times 69 \times 137} 6 - \frac {21 \times 22 \times 43} 6\)
\(\ds \) \(=\) \(\ds 34 \times 23 \times 137 - 7 \times 11 \times 43\)
\(\ds \) \(=\) \(\ds 107 \, 134 - 3311\)
\(\ds \) \(=\) \(\ds 103 \, 823\)
\(\ds \) \(=\) \(\ds 47^3\)

$\blacksquare$


Sources

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