Prove:1/1.5+1/5.9+1/9.13+...+1/(4n-1) (4n+1)=n/4n+1
We can prove this statement by using mathematical induction.
Base case:
n = 1
1/1.5 + 1/5.9 = 1/3.5 + 1/5.9 = 0.2857142857 (approximately)
1/(4 * 1 - 1)(4 * 1 + 1) = 1/3(5) = 1/15 = 0.0666666667 (approximately)
We can see that both sides are not equal when n = 1.
Inductive step:
Let's assume that the equation is true for n = k:
1/1.5 + 1/5.9 + 1/9.13 + … + 1/(4k-1)(4k+1) = k/4k+1
We need to prove that this equation is also true for n = k + 1:
1/1.5 + 1/5.9 + 1/9.13 + … + 1/(4k-1)(4k+1) + 1/(4k+3)(4k+5) = (k+1)/4(k+1) + 1
Now, we can rewrite the equation that we assumed is true for n = k:
k/4k+1 + 1/(4k+3)(4k+5)
Combining these two equations, we get:
k/4k+1 + 1/(4k+3)(4k+5) = k/4k+1 + (4k+1)/(4k+1)(4k+5) = (k(4k+1) + 1) / (4k+1)(4k+5) = (4k^2 + k + 1) / (4k^2 + 6k + 5) = (4k^2 + 6k + 4 + k + 1) / (4k^2 + 6k + 5) = ((2k + 2)(2k + 2) + (k+1)) / (4k + 5)(k + 1) = (2k + 2)(2k + 2) + (k+1)
This shows that the equation is also true for n = k + 1.
Therefore, by mathematical induction, we have proven the given statement.