Show that
\[\sum_{j=1}^n j =\frac{1}{2}n(n+1)\]
Hint: notice that \(1+n=n+1\), \(2+(n-1)=n+1\), \(3+(n-2)=n+1\) and so on. How many such pairings exist ?
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Home » Prerequisite » Series » Arithmetic sum
Show that
\[\sum_{j=1}^n j =\frac{1}{2}n(n+1)\]
Hint: notice that \(1+n=n+1\), \(2+(n-1)=n+1\), \(3+(n-2)=n+1\) and so on. How many such pairings exist ?
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