Sum of Sequence of Squares/Proof by Telescoping Series

Theorem

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

Proof

Observe that:

$3 i \paren {i + 1} = i \paren {i + 1} \paren {i + 2} - i \paren {i + 1} \paren {i - 1}$

That is:

$(1): \quad 6 T_i = \paren {i + 1} \paren {\paren {i + 1} + 1} \paren {\paren {i + 1} - 1} - i \paren {i + 1} \paren {i - 1}$

Then:

\(\ds n^2\)

\(=\)

\(\ds \frac {n^2 + n + n^2 - n} 2\)

\(\ds \)

\(=\)

\(\ds \frac {n \paren {n + 1} + n \paren {n - 1} } 2\)

\(\ds \)

\(=\)

\(\ds \frac {n \paren {n + 1} } 2 + \frac {n \paren {n - 1} } 2\)

\(\ds \)

\(=\)

\(\ds T_n + T_{n-1}\)

where $T_n$ is the $n$th Triangular number.

Then:

\(\ds \sum_{i \mathop = 1}^n i^2\)

\(=\)

\(\ds 1 + \paren {T_1 + T_2} + \paren {T_2 + T_3} + \paren {T_3 + T_4} + \cdots + \paren {T_{n - 1} + T_n}\)

\(\ds \)

\(=\)

\(\ds 1 + 2 T_2 + 2 T_3 + 2 T_4 + \cdots + 2 T_{n - 1} + T_n\)

\(\ds \)

\(=\)

\(\ds 1 - T_1 - T_n + 2 \paren {T_1 + T_2 + T_3 + T_4 + \cdots + T_n}\)

\(\ds \)

\(=\)

\(\ds 2 \paren {\sum_{i \mathop = 1}^n T_i} - \frac {n \paren {n + 1} } 2\)

\(\ds \)

\(=\)

\(\ds 2 \paren {\frac {n \paren {n + 1} \paren {n + 2} } 6} - \frac {n \paren {n + 1} } 2\)

Telescoping Series from $(1)$ above

\(\ds \)

\(=\)

\(\ds \frac {2 n \paren {n^2 + 3 n + 2} - \paren {3 n^2 + 3 n} } 6\)

\(\ds \)

\(=\)

\(\ds \frac {2 n^3 + 3 n^2 + n} 6\)

\(\ds \)

\(=\)

\(\ds \frac {n \paren {n + 1} \paren {2 n + 1} } 6\)

$\blacksquare$

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