Sums of Sequences of Consecutive Squares which are Square

From ProofWiki
Jump to navigation Jump to search

Theorem

The sums of the following sequences of successive squares are themselves square:

\(\ds \sum_{i \mathop = 7}^{29} k^2\) \(=\) \(\ds 7^2 + 8^2 + \cdots + 29^2\)
\(\ds \sum_{i \mathop = 7}^{39} k^2\) \(=\) \(\ds 7^2 + 8^2 + \cdots + 39^2\)
\(\ds \sum_{i \mathop = 7}^{56} k^2\) \(=\) \(\ds 7^2 + 8^2 + \cdots + 56^2\)
\(\ds \sum_{i \mathop = 7}^{190} k^2\) \(=\) \(\ds 7^2 + 8^2 + \cdots + 190^2\)


Proof

From Sum of Sequence of Squares:

$\ds \forall n \in \N: \sum_{i \mathop = 1}^n i^2 = \frac {n \paren {n + 1} \paren {2 n + 1} } 6$


Thus:

\(\ds \sum_{i \mathop = 7}^{29} i^2\) \(=\) \(\ds \sum_{i \mathop = 1}^{29} i^2 - \sum_{i \mathop = 1}^6 i^2\)
\(\ds \) \(=\) \(\ds \frac {29 \left({29 + 1}\right) \left({2 \times 29 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6\)
\(\ds \) \(=\) \(\ds \frac {29 \times 30 \times 59} 6 - \frac {6 \times 7 \times 13} 6\)
\(\ds \) \(=\) \(\ds 8 \, 555 - 91\)
\(\ds \) \(=\) \(\ds 8 \, 464\)
\(\ds \) \(=\) \(\ds 92^2\)


\(\ds \sum_{i \mathop = 7}^{39} i^2\) \(=\) \(\ds \sum_{i \mathop = 1}^{39} i^2 - \sum_{i \mathop = 1}^6 i^2\)
\(\ds \) \(=\) \(\ds \frac {39 \left({39 + 1}\right) \left({2 \times 39 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6\)
\(\ds \) \(=\) \(\ds \frac {39 \times 40 \times 79} 6 - \frac {6 \times 7 \times 13} 6\)
\(\ds \) \(=\) \(\ds 20 \, 540 - 91\)
\(\ds \) \(=\) \(\ds 20 \, 449\)
\(\ds \) \(=\) \(\ds 143^2\)


\(\ds \sum_{i \mathop = 7}^{56} i^2\) \(=\) \(\ds \sum_{i \mathop = 1}^{56} i^2 - \sum_{i \mathop = 1}^6 i^2\)
\(\ds \) \(=\) \(\ds \frac {56 \left({56 + 1}\right) \left({2 \times 56 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6\)
\(\ds \) \(=\) \(\ds \frac {56 \times 57 \times 113} 6 - \frac {6 \times 7 \times 13} 6\)
\(\ds \) \(=\) \(\ds 60 \, 116 - 91\)
\(\ds \) \(=\) \(\ds 60 \, 025\)
\(\ds \) \(=\) \(\ds 245^2\)


\(\ds \sum_{i \mathop = 7}^{190} i^2\) \(=\) \(\ds \sum_{i \mathop = 1}^{190} i^2 - \sum_{i \mathop = 1}^6 i^2\)
\(\ds \) \(=\) \(\ds \frac {190 \left({190 + 1}\right) \left({2 \times 190 + 1}\right)} 6 - \frac {6 \left({6 + 1}\right) \left({2 \times 6 + 1}\right)} 6\)
\(\ds \) \(=\) \(\ds \frac {190 \times 191 \times 381} 6 - \frac {6 \times 7 \times 13} 6\)
\(\ds \) \(=\) \(\ds 2 \, 304 \, 415 - 91\)
\(\ds \) \(=\) \(\ds 2 \, 304 \, 324\)
\(\ds \) \(=\) \(\ds 1 \, 518^2\)

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Sums_of_Sequences_of_Consecutive_Squares_which_are_Square&oldid=287054"