Discrete mathematics
Mathematical induction
Prove:1/1.5+1/5.9+1/9.13+....+1/(4n-3)(4n+1)=n/4n+1
Let P(n) be the statement: 1/1.5 + 1/5.9 + 1/9.13 + ... + 1/(4n-3)(4n+1) = n/(4n+1).
Base Case:
For n = 1, the left side of the equation becomes 1/1.5 = 1/3 and the right side is 1/(4*1+1) = 1/5. These are equal, so the statement holds true for n = 1.
Inductive Step:
Assume the statement P(k) is true for some arbitrary positive integer k, where k >= 1.
So we have: 1/1.5 + 1/5.9 + 1/9.13 + ... + 1/(4k-3)(4k+1) = k/(4k+1).
Now consider P(k+1):
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/(4k+1) + 1/(4k+1+4)(4k+1+1)
= k/(4k+1) + 1/(4(k+1)+1)
= k/(4k+1) + (k+1)/(4(k+1)+1)
= [k(4(k+1)+1) + (4k+1)(k+1)] / [(4k+1)(4(k+1)+1)]
= [4k(k+1) + k + 4k(k+1) + 1] / [16k(k+1) + 1]
= [8k(k+1) + 1] / [16k(k+1) + 1]
= (k+1) / (4(k+1) + 1).
Therefore, P(k+1) is true if P(k) is true. By the principle of mathematical induction, P(n) is true for all positive integers n >= 1. Thus, the given statement is proven.