Difference between revisions of "T(n+1)=T(n)+n+1"
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==Theorem== | ==Theorem== | ||
− | The following formula holds: | + | The following formula holds for $n=1,2,3,\ldots$: |
$$T(n+1) = T(n)+n+1,$$ | $$T(n+1) = T(n)+n+1,$$ | ||
where $T(n)$ denotes the $n$th [[triangular numbers|triangular number]]. | where $T(n)$ denotes the $n$th [[triangular numbers|triangular number]]. | ||
Line 7: | Line 7: | ||
==References== | ==References== | ||
− | * {{PaperReference|Triangular numbers|1974|V.E. Hoggatt, Jr|author2=Marjorie Bicknell|prev=T(n)=n(n+1)/2|next= | + | * {{PaperReference|Triangular numbers|1974|V.E. Hoggatt, Jr|author2=Marjorie Bicknell|prev=T(n)=n(n+1)/2|next=n^2=T(n)+T(n-1)}} $(1.2)$ |
[[Category:Theorem]] | [[Category:Theorem]] | ||
[[Category:Unproven]] | [[Category:Unproven]] |
Latest revision as of 01:28, 30 May 2017
Theorem
The following formula holds for $n=1,2,3,\ldots$: $$T(n+1) = T(n)+n+1,$$ where $T(n)$ denotes the $n$th triangular number.
Proof
References
- V.E. Hoggatt, Jr and Marjorie Bicknell: Triangular numbers (1974)... (previous)... (next) $(1.2)$