Discrete math (mathematical induction)
Prove:1/1.5+1/5.9+1/9.13+....+1/(4n-3) (4n+1)=n/4n+1
To prove the given statement using mathematical induction, we need to show that it holds for the base case and then prove that if it holds for n, it also holds for n+1.
Base Case: n = 1
When n = 1, the expression becomes 1/1.5 = 1/4 = 1/(4*1+1) = 1/5. Hence, the base case holds.
Inductive Step:
Assume that the statement holds for some arbitrary positive integer k, i.e.,
1/1.5 + 1/5.9 + 1/9.13 + .... + 1/(4k-3)(4k+1) = k/(4k+1)
Now, we need to prove that the statement also holds for k+1, i.e.,
1/1.5 + 1/5.9 + 1/9.13 + .... + 1/(4k-3)(4k+1) + 1/(4(k+1)-3)(4(k+1)+1) = (k+1)/(4(k+1)+1)
Adding the next term in the sequence:
1/1.5 + 1/5.9 + 1/9.13 + .... + 1/(4k-3)(4k+1) + 1/(4(k+1)-3)(4(k+1)+1)
= k/(4k+1) + 1/(4(k+1)-3)(4(k+1)+1)
= (k(16(k+1) + 4k - 3) + 1)/(16(k+1) + 1)
= (16k^2 + 20k + 1)/(16k + 5)
Now, we need to prove that this expression equals (k+1)/(4(k+1)+1).
Compare the two expressions to show that they are equal by simplification:
16k^2 + 20k + 1 = 4k^2 + 5k + 1
16k + 5 = 4k + 1
This shows that the statement holds for n+1 as well.
Since the base case holds and the inductive step proves that if the statement holds for n, it also holds for n+1, by the principle of mathematical induction, the original statement is proved.