How do you find the closed form of summations such as:
$$\sum\limits_{k=1}^{n+1} k$$
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1Are you asking what it means, or are you asking for a closed form? – Brian M. Scott Dec 09 '13 at 00:20
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$1+2+3+\cdots+n+n+1$ – Jemmy Dec 09 '13 at 00:21
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1be Gauss and about age 6. – James S. Cook Dec 09 '13 at 00:21
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Oh, yes I am wondering about the closed form. – camelbak Dec 09 '13 at 00:26
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$$\displaystyle\sum\limits_{k=1}^{n+1} k = \left( \sum\limits_{k=1}^{n} k \right) + (n + 1)$$ because $n+1$ is the last term.
Xoque55
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mathematical notation is like a language and its correct use depends upon knowledge of syntax and conventions. the expression: $$ \sum_{k=1}^{n+1} f(k) $$ means $$ f(1)+f(2)+f(3)+\dots +f(n)+f(n+1) $$ a similar expression $$ \sum_{k=1}^{n+\frac12} f(k) $$ is recognized as meaningless because of unstated conventions which govern the use of the symbols. look at the grammatical convention which makes the following a true statement: $$ 1 = \frac12 + \frac14 + \frac18+ \dots $$ it would be inappropriate to conclude from this that $$ \dots=\frac18 $$ since the same symbol $\dots$ has a meaning which is determined by the context.
if i have misunderstood what you are puzzled about, and you simply want to evaluate the particular expression, then note that if $$ S_f(n) = \sum_{k=1}^n f(k) $$ then necessarily (although this requires a little thought) you must also have
$$ S_f(n) = \sum_{k=1}^n f(n+1-k) $$ from this follows, by addition: $$ 2S_f(n) = \sum_{k=1}^n [f(k)+f(n+1-k)] $$ in the particularly simple case where $f(k)=k$ this reduces to $$ 2S_f(n) = \sum_{k=1}^n [k+n+1-k] = \sum_{k=1}^n [n+1] $$ since $n+1$ is independent of $k$ then we are simply summing $n$ terms, each of which is $n+1$ so the sum must be $n(n+1)$, and for this particular $f$ we have: $$ 2S_f(n) = n(n+1) $$ hence: $$ S_f(n) = \frac12 n(n+1) $$ which is the $n^{th}$ triangular number
David Holden
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