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Sagot :
Sure! Let's start by looking at the left-hand side of the given equation:
[tex]\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)} \][/tex]
We need to show that this sum equals:
[tex]\[ \frac{n}{2n + 1} \][/tex]
Step-by-Step Solution:
1. Consider a general term:
The general term in the sum is:
[tex]\[ \frac{1}{(2k-1)(2k+1)} \][/tex]
for [tex]\( k \)[/tex], ranging from 1 to [tex]\( n \)[/tex].
2. Simplify the general term:
Notice that:
[tex]\[ \frac{1}{(2k-1)(2k+1)} \][/tex]
can be simplified using partial fractions. We decompose it as follows:
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1} \][/tex]
To find [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we set up the equation:
[tex]\[ 1 = A(2k+1) + B(2k-1) \][/tex]
Expanding and combining terms, we get:
[tex]\[ 1 = (2A + 2B)k + (A - B) \][/tex]
For the equation to hold for all [tex]\( k \)[/tex], the coefficients of [tex]\( k \)[/tex] and the constant term must both match:
[tex]\[ 2A + 2B = 0 \quad \text{and} \quad A - B = 1 \][/tex]
Solving these simultaneous equations, we find:
[tex]\[ 2A + 2B = 0 \Rightarrow A + B = 0 \Rightarrow B = -A \][/tex]
[tex]\[ A - B = 1 \Rightarrow A - (-A) = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2} \][/tex]
Therefore, [tex]\( B = -\frac{1}{2} \)[/tex].
So, our partial fractions decomposition is:
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{1/2}{2k-1} - \frac{1/2}{2k+1} \][/tex]
Therefore,
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \][/tex]
3. Sum the series:
Using the simplified form, the sum becomes:
[tex]\[ \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \][/tex]
Notice that this is a telescoping series. Most terms will cancel out:
[tex]\[ \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{(2n-1)} - \frac{1}{(2n+1)} \right) \right) \][/tex]
The intermediate terms cancel, leaving:
[tex]\[ \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) \][/tex]
Simplifying this, we get:
[tex]\[ \frac{1}{2} \left( \frac{2n+1 - 1}{2n+1} \right) = \frac{1}{2} \left( \frac{2n}{2n+1} \right) = \frac{n}{2n+1} \][/tex]
Thus, we have shown that:
[tex]\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1} \][/tex]
This completes the proof.
[tex]\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)} \][/tex]
We need to show that this sum equals:
[tex]\[ \frac{n}{2n + 1} \][/tex]
Step-by-Step Solution:
1. Consider a general term:
The general term in the sum is:
[tex]\[ \frac{1}{(2k-1)(2k+1)} \][/tex]
for [tex]\( k \)[/tex], ranging from 1 to [tex]\( n \)[/tex].
2. Simplify the general term:
Notice that:
[tex]\[ \frac{1}{(2k-1)(2k+1)} \][/tex]
can be simplified using partial fractions. We decompose it as follows:
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1} \][/tex]
To find [tex]\( A \)[/tex] and [tex]\( B \)[/tex], we set up the equation:
[tex]\[ 1 = A(2k+1) + B(2k-1) \][/tex]
Expanding and combining terms, we get:
[tex]\[ 1 = (2A + 2B)k + (A - B) \][/tex]
For the equation to hold for all [tex]\( k \)[/tex], the coefficients of [tex]\( k \)[/tex] and the constant term must both match:
[tex]\[ 2A + 2B = 0 \quad \text{and} \quad A - B = 1 \][/tex]
Solving these simultaneous equations, we find:
[tex]\[ 2A + 2B = 0 \Rightarrow A + B = 0 \Rightarrow B = -A \][/tex]
[tex]\[ A - B = 1 \Rightarrow A - (-A) = 1 \Rightarrow 2A = 1 \Rightarrow A = \frac{1}{2} \][/tex]
Therefore, [tex]\( B = -\frac{1}{2} \)[/tex].
So, our partial fractions decomposition is:
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{1/2}{2k-1} - \frac{1/2}{2k+1} \][/tex]
Therefore,
[tex]\[ \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \][/tex]
3. Sum the series:
Using the simplified form, the sum becomes:
[tex]\[ \sum_{k=1}^n \frac{1}{(2k-1)(2k+1)} = \frac{1}{2} \sum_{k=1}^n \left( \frac{1}{2k-1} - \frac{1}{2k+1} \right) \][/tex]
Notice that this is a telescoping series. Most terms will cancel out:
[tex]\[ \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{5} \right) + \cdots + \left( \frac{1}{(2n-1)} - \frac{1}{(2n+1)} \right) \right) \][/tex]
The intermediate terms cancel, leaving:
[tex]\[ \frac{1}{2} \left( 1 - \frac{1}{2n+1} \right) \][/tex]
Simplifying this, we get:
[tex]\[ \frac{1}{2} \left( \frac{2n+1 - 1}{2n+1} \right) = \frac{1}{2} \left( \frac{2n}{2n+1} \right) = \frac{n}{2n+1} \][/tex]
Thus, we have shown that:
[tex]\[ \frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1} \][/tex]
This completes the proof.
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