- #1
thefeedinghan
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Hello there, I'm having trouble proving this by induction
\frac{1}{1(2)} + 1\frac{1}{2(3)}+...+\frac{1}{n(n+1)} = \frac{n}{n+1}
For the base case n=1
\frac{1}{1(2)}=\frac{1}{1+1} = \frac{1}{2} = \frac{1}{2}
\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} <- the second term would be the next integer
\frac{1}{1(2)}+ \frac{1}{2(3)}+\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{(k+1)+1} = \frac{k}{k+2} + \frac{1}{k+2}
I don't know where to go from here, any help would be appreciated
Homework Statement
\frac{1}{1(2)} + 1\frac{1}{2(3)}+...+\frac{1}{n(n+1)} = \frac{n}{n+1}
Homework Equations
For the base case n=1
\frac{1}{1(2)}=\frac{1}{1+1} = \frac{1}{2} = \frac{1}{2}
\frac{k}{k+1} + \frac{1}{(k+1)(k+2)} <- the second term would be the next integer
The Attempt at a Solution
\frac{1}{1(2)}+ \frac{1}{2(3)}+\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{(k+1)+1} = \frac{k}{k+2} + \frac{1}{k+2}
I don't know where to go from here, any help would be appreciated
