## [2^2 + 4^2 + 6^2 +…+ (2n)^2] – [1^2 + 3^2 + 5^2 +…+ (2n-1)^2]

An integer of the form   $T_n \; = \; n \,(n + 1)/2$   is called a Triangular number.

$T_2 \; = \; 3 \; = \; 2^2 \; - \; 1^2$

$T_4 \; = \; 10 = (2^2 + 4^2) \; - \; (1^2 + 3^2)$

$T_6 \; = \; 21 = (2^2 + 4^2 + 6^2) \; - \; (1^2 + 3^2 + 5^2)$

$T_8 \; = \; 36 = (2^2 + 4^2 + 6^2 + 8^2) \; - \; (1^2 + 3^2 + 5^2 + 7^2)$

$T_{10} \; = \; 55 \; = \; (2^2 + 4^2 + ... + 10^2) \; - \; (1^2 + 3^2 + ... + 11^2)$

$A \; = \; 2^2 + 4^2 + 6^2 +... + (2 \, n)^2 \; = \; 2/3 \; n \, (n+1) \, (2 n+1)$

$B \; = \; 1^2 + 3^2 + 5^2 + ... + (2 \, n-1)^2 \; = \; 1/3 \; (4 n^3-n)$

$A \; - \; B$
$= 2/3 \; n \, (n+1) \, (2 \, n+1) \; - \; 1/3 \; (4 \, n^3-n)$
$= n \, (2 \, n+1)$
$= T_{ \,2 \,n \,}$

$D \; = \; 1^2 \; + \; 3^2 \; + \; 5^2 \; + \; ... \; + \; (2 \, n+1)^2 \; = \; 1/3 \; (n+1) \, (2 \, n+1) \, (2 \, n+3)$

$A \; = \; 2^2 \; + \; 4^2 \; + \; 6^2 \; + \; ... \; + (2 \, n)^2 \; = \; 2/3 \; n \, (n+1) \, (2 \, n+1)$

$D \; - \; A$
$= \; 1/3 \; (n+1) \, (2 \, n+1) \, (2 \, n+3) \; - \; 2/3 \; n \, (n+1) \, (2 \, n+1)$
$= \; (2 \, n+1) \, (n+1)$