prove by mathematical induction that 1/n(n+1) +1/(n+1)(n+2) +1/(n+2)(n+3) +⋯+1/(2n-1)2n=1/2n and please show me the step by step solutions

To prove the statement
\[
\frac{1}{n(n+1)} + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)} + \cdots + \frac{1}{(2n-1)2n} = \frac{1}{2n}
\]
we’ll use **mathematical induction**.

### Step 1: Base Case (\(n = 1\))
For \(n = 1\), the left side of the equation is:
\[
\frac{1}{1 \cdot 2} = \frac{1}{2}
\]
The right side of the equation is also:
\[
\frac{1}{2 \cdot 1} = \frac{1}{2}
\]
So, the base case holds.

### Step 2: Inductive Hypothesis
Assume that the statement is true for some integer \(n = k\), i.e.,
\[
\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(2k-1)2k} = \frac{1}{2k}
\]

### Step 3: Inductive Step
We need to show that the statement holds for \(n = k + 1\), i.e.,
\[
\frac{1}{(k+1)(k+2)} + \frac{1}{(k+2)(k+3)} + \cdots + \frac{1}{(2k+1)(2k+2)} = \frac{1}{2(k+1)}
\]
Starting with the left side of the expression for \(n = k + 1\), we add the next term \(\frac{1}{(2k+1)(2k+2)}\) to both sides of the inductive hypothesis:
\[
\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} + \cdots + \frac{1}{(2k-1)2k} + \frac{1}{(2k+1)(2k+2)}
\]
Using the inductive hypothesis, we replace the sum up to \(\frac{1}{(2k-1)2k}\) with \(\frac{1}{2k}\):
\[
\frac{1}{2k} + \frac{1}{(2k+1)(2k+2)}
\]

#### Combining the Terms
Now we combine the two terms on the right side:
\[
\frac{1}{2k} + \frac{1}{(2k+1)(2k+2)} = \frac{(2k+2) + 1}{2k(2k+2)}
\]
Simplify the numerator:
\[
= \frac{2k+3}{2k(2k+2)}
\]


Prove by mathematical induction that 1/n(n+1) + 1/(n+1)(n+2) + 1/(n+2)(n+3) + ⋯ + 1/(2n-1)2n = 1/2n and please show me the step-by-step solutions.

This step-by-step guide explains how to use mathematical induction to prove the series summation from 1/n(n+1) to 1/2n. Learn the base case and inductive step, with clear explanations suitable for high school students (Grades 11-12).

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