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OpenStudy (anonymous):

Can someone explain why the following is correct for n=1:

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OpenStudy (anonymous):

\[\sum_{k=2}^{2n+1} 1/k(k+1) = n/2(n+1)\]

OpenStudy (anonymous):

For \(n=1\), we have \(\sum_{k=2}^{3}\frac{1}{k(k+1)}=\frac{1}{2(3)}+\frac{1}{3(4)}=\frac{2+1}{3(4)}=\frac{1}{4}=\frac{n}{2(n+1)} \text{ (since n=1)}\).

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