Some Triangular number identities — Part 1

 
 
Prove the following;

 
T_{3 \, n+1} = 9 \, T_{n} + 1

T_{7 \, n+3} = 49 \, T_{n} + 6

T_{5 \, n+1} = T_{3 \, n} + T_{4 \, n+1}

T_{5 \, n+3} = T_{4 \, n+2} + T_{3 \, n+2}

T_{13 \, n+10} = T_{5 \, n+4} + T_{12 \, n+9}

T_{17 \, n+10} = T_{8 \, n+4} + T_{15 \, n+9}

T_{3 \, n - 1} = 2 \, T_{2 \, n - 1} + T_{n}

T_{3 \, n} = 2 \, T_{2 \, n} + T_{n - 1}

T_{3 \, n + 1} = T_{2 \, n} + T_{2 \, n + 1} + T_{n}

3 \, T_{n} + 1 = T_{n - 1} + T_{n} + T_{n + 1}

T_{4 \, n^2 + 5 \, n + 2} = T_{4 \, n^2 + 5 \, n} + T_{4 \, n + 2}

T_{m + n} = T_{m} + T_{n} + m \, n

T_{m + n + 1} = T_{m} + T_{n} + (m + 1) \,(n + 1)

T_{m + n - 1} = T_{m} + T_{n} + m \, n - m - n

T_{n - m} = T_{m} + T_{n} - m \,(n + 1)

T_{n - m - 1} = T_{m} + T_{n} - n \,(m + 1)

T_{m \, n} = T_{m} \, T_{n} + T_{n-1} \, T_{m-1}

T_{m \, n + 1} = T_{m} \, T_{n} + T_{n-1} \, T_{m-1} + m \, n + 1

T_{m \, n - 1} = T_{m} \, T_{n} + T_{n-1} \, T_{m-1} - m \, n

T_{n^2} = ( \,T_{n} \,)^2 + ( \,T_{n-1} \,)^2

T_{n^2 + m^2} = ( \,T_{n} \,)^2 + ( \,T_{n-1} \,)^2 + ( \,T_{m} \,)^2 + ( \,T_{m-1} \,)^2 - m^2 \, n^2

T_{n^2 - m^2} = ( \,T_{n} \,)^2 + ( \,T_{n-1} \,)^2 + ( \,T_{m} \,)^2 + ( \,T_{m-1} \,)^2 - m^2 \,(n^2+1)

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

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