Discrete mathematics
Mathematical induction
Prove: 1/1.5+1/5.9+1/9.3+....+1/(4n-3)(4n+1)=n/4n+1
To prove this statement by mathematical induction, we will follow these steps:
1. Show that the statement is true for n = 1:
For n = 1, we have:
1/1.5 = 1/(4(1)-3)(4(1)+1) = 1/5, and n/(4n+1) = 1/(4*1+1) = 1/5
Therefore, the statement is true for n = 1.
2. Assume that the statement is true for some positive integer k:
Assume that:
1/1.5 + 1/5.9 + 1/9.3 + ... + 1/(4k-3)(4k+1) = k/4k+1
3. Show that the statement is true for n = k + 1:
We need to prove that:
1/1.5 + 1/5.9 + 1/9.3 + ... + 1/(4k-3)(4k+1) + 1/(4(k+1)-3)(4(k+1)+1) = (k+1)/4(k+1)+1
Using the assumption in step 2, we have:
k/4k+1 + 1/(4(k+1)-3)(4(k+1)+1) = k/4k+1 + 1/(4k+4-3)(4k+4+1) = k/4k+1 + 1/(4k+1)(4k+5)
Now, we need to find a common denominator for the two fractions:
k(4k+5)/(4k+1)(4k+5) + 1/(4k+1)(4k+5) = (k(4k+5) + 1)/(4k+1)(4k+5) = (4k^2 + 5k + 1)/(16k^2 + 24k + 5)
Since (4k^2 + 5k + 1)/(16k^2 + 24k + 5) = (k+1)/(4k+1), the statement is true for n = k + 1.
By mathematical induction, the statement is true for all positive integers n.