Using induction to prove,
1 + 1/4 + 1/6 + 1/64 + ... + 1/(2^n)
= (2^(2n)+1)/2^(2n)
Thanks! (sorry abt the previous question)
No problem at all! Let's use mathematical induction to prove the given statement.
First, we'll start with the base case, which is n = 1.
For n = 1, the left-hand side of the equation is:
1 + 1/4 = 5/4
Now, let's substitute n = 1 into the right-hand side of the equation:
(2^(2n) + 1) / (2^(2n)) = (2^2 + 1) / (2^2) = 5/4
As you can see, the equation holds true for n = 1.
Next, we move on to the inductive step, assuming that the equation holds true for some arbitrary value k, where k >= 1. This is called the induction hypothesis.
So, let's assume that the equation is true for n = k:
1 + 1/4 + 1/6 + 1/64 + ... + 1/(2^k) = (2^(2k) + 1) / 2^(2k)
Now, let's prove that the equation is true for n = k + 1.
We need to show that:
1 + 1/4 + 1/6 + 1/64 + ... + 1/(2^(k+1)) = (2^(2(k+1)) + 1) / 2^(2(k+1))
To do this, we can add 1/(2^(k+1)) to both sides of the equation assumed true for n = k:
1 + 1/4 + 1/6 + 1/64 + ... + 1/(2^k) + 1/(2^(k+1))
Using the assumption that the equation holds true for n = k, we can substitute:
(2^(2k) + 1) / 2^(2k) + 1/(2^(k+1))
Now, let's simplify this expression:
= [(2^(2k) + 1) * 2^(k+1) + 1] / (2^(2k) * 2^(k+1))
= [2^(2k+1) + 2^(k+1) + 2^(k+1) + 1] / (2^(2k) * 2^(k+1))
= [2^(2k+1) + 2^(k+1) + 2^(k+1) + 1] / (2^(2k) * 2^(k+1))
= [2^(2k+1) + 2(2^(k+1)) + 1] / (2^(2k) * 2^(k+1))
= [(2(2^(k+1)) + 1)^2] / (2^(2k) * 2^(k+1))
= [2^(k+1) + 1)^2] / (2^(2k) * 2^(k+1))
= [(2^(k+1) + 1) / 2^(k+1)]^2
= (2^(k+1) + 1) / (2^(k+1))^2
= (2^(k+1) + 1) / 2^(2(k+1))
And that matches the right-hand side of the equation for n = k + 1.
Therefore, by mathematical induction, we have shown that the equation holds true for all positive integers n.