solve this using the mathematical induction : 1/1(3)+1/3(5)+1/5(7)+......+1/(2n-1)(2n+1)=n/(2n+1)

To prove this equation using mathematical induction, we will follow the three steps: base case, inductive hypothesis, and inductive step.

Step 1: Base Case
Let's check if the equation holds for the initial value, which is n = 1:

1/(1*3) = 1/(2*1+1)
1/3 = 1/3

The equation holds true for n = 1.

Step 2: Inductive Hypothesis
Assume that the equation holds true for some positive integer k, which means:

1/1(3) + 1/3(5) + 1/5(7) + ... + 1/(2k-1)(2k+1) = k/(2k+1)

Step 3: Inductive Step
We need to prove that the equation will hold true for the next positive integer, which is k + 1:

1/1(3) + 1/3(5) + 1/5(7) + ... + 1/(2k-1)(2k+1) + 1/(2k+1)(2k+3) = (k+1)/(2(k+1)+1)

By assuming the inductive hypothesis, we can rewrite the left-hand side of the equation:

k/(2k+1) + 1/(2k+1)(2k+3) = (k+1)/(2(k+1)+1)

Now, let's simplify each side of the equation individually:

Left Side:
k/(2k+1) + 1/(2k+1)(2k+3) =
(k(2k+3) + 1) / (2k+1)(2k+3) =
(2k^2 + 3k + 1) / (2k+1)(2k+3)

Right Side:
(k+1)/(2(k+1)+1) =
(k+1)/(2k+2+1) =
(k+1)/(2k+3)

The left and right sides of the equation are equal, so the equation holds true for k + 1.

By completing these steps, we have proved that the equation holds true for all positive integers using mathematical induction.

To solve this using mathematical induction, we will first establish the base case and then prove the induction step.

Step 1: Base Case
Let n = 1:
1/(1(3)) = 1/(2+1) = 1/3
Therefore, the equation holds for n = 1.

Step 2: Induction Hypothesis
Assume that the equation holds for some arbitrary positive integer k:
1/1(3) + 1/3(5) + 1/5(7) + ... + 1/(2k-1)(2k+1) = k/(2k+1)

Step 3: Induction Step
We must now prove that the equation also holds for k + 1.

By assuming the equation holds for k and adding 1/(2k+1)(2k+3) to both sides of the equation, we get:
1/1(3) + 1/3(5) + 1/5(7) + ... + 1/(2k-1)(2k+1) + 1/(2k+1)(2k+3) = k/(2k+1) + 1/(2k+1)(2k+3)

Next, we simplify the expression on the right-hand side:
k/(2k+1) + 1/(2k+1)(2k+3) = ((k)(2k+3) + 1)/(2k+1)(2k+3) = (2k^2 + 3k + 1)/(2k+1)(2k+3)

Our goal is to show that the sum of the left-hand side above is equal to (k + 1)/(2k + 3).

Notice that: 2k^2 + 3k + 1 = (2k^2 + 4k + 2) - (k + 1) = 2(k + 1)^2 - (k + 1)

Now we can substitute this expression back into our equation:
(2(k + 1)^2 - (k + 1))/((2k+1)(2k+3)) = (k + 1)/(2k + 3)

Therefore, the equation holds for k + 1.

By the principle of mathematical induction, we have proven that the given equation holds for all positive integers n.

First, verify for n=1:

1/1*3 = 1/(3) -- true

Assume it is true for n:

1/1*3 + 1/3*5 + ... + 1/(2n-1)(2n+1) = n/(2n+1)

Now, add the next term to both sides:

1/1*3 + 1/3*5 + ... + 1/(2n-1)(2n+1) + 1/(2n+1)(2n+3) = n/(2n+1) + 1/(2n+1)(2n+3)

= [n(2n+3) + 1]/(2n+1)(2n+3)
= (2n^2 + 3n + 1)/(2n+1)(2n+3)
= (2n+1)(n+1)/(2n+1)(2n+3)
= (n+1)/(2(n+1)+1)

This is the formula, with n replaced by n+1.

So, if it's true for n, it's true for n+1.
It's true for n=1, so therefore also for n=2,3,4,...