To determine which expression is obtained by induction at [tex]\( n = k + 1 \)[/tex], we should start by looking at the sum [tex]\(\sum \frac{1}{r(r+1)}\)[/tex].
By the method of partial fractions, we know that:
[tex]\[
\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}
\][/tex]
This allows us to write the sum [tex]\( \sum_{r=1}^n \frac{1}{r(r+1)} \)[/tex] as a telescoping series:
[tex]\[
\sum_{r=1}^n \left( \frac{1}{r} - \frac{1}{r+1} \right)
\][/tex]
When we expand this sum, the intermediate terms cancel out:
[tex]\[
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{n} - \frac{1}{n+1} \right)
\][/tex]
All terms except the first and the last cancel, leaving us with:
[tex]\[
1 - \frac{1}{n+1} = \frac{n}{n+1}
\][/tex]
Now, using mathematical induction, suppose the given expression holds true for [tex]\( n = k \)[/tex]:
[tex]\[
\sum_{r=1}^k \frac{1}{r(r+1)} = \frac{k}{k+1}
\][/tex]
We need to prove that the expression also holds for [tex]\( n = k + 1 \)[/tex]. Consider:
[tex]\[
\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \left( \sum_{r=1}^k \frac{1}{r(r+1)} \right) + \frac{1}{(k+1)((k+1)+1)}
\][/tex]
From the induction hypothesis, we know:
[tex]\[
\sum_{r=1}^k \frac{1}{r(r+1)} = \frac{k}{k+1}
\][/tex]
Thus,
[tex]\[
\sum_{r=1}^{k+1} \frac{1}{r(r+1)} = \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}
\][/tex]
We need to simplify this expression:
[tex]\[
\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}
\][/tex]
Combining these fractions over the common denominator [tex]\((k+1)(k+2)\)[/tex], we get:
[tex]\[
\frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}
\][/tex]
Simplifying the fraction, we get:
[tex]\[
\frac{k+1}{k+2}
\][/tex]
Therefore, by induction, the simplified form at [tex]\( n = k+1 \)[/tex] is:
[tex]\[
\frac{k+1}{k+2}
\][/tex]
So the correct choice is:
(c) [tex]\(\frac{k+1}{k+2}\)[/tex]
