[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)

 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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