Observe that :
1/1*3= 1/3
1/1*3+ 1/3*5= 2/5
1/1*3+ 1/3*5+ 1/5*7= 3/7
1/1*3+ 1/3*5+ 1/5*7+ 1/7*9= 4/9
Come up with a formula s(n) for the above, for n is greater than or equal to 1. Prove it by mathematical induction
To come up with a formula s(n) for the given pattern, let's first analyze the pattern that emerges:
1/1 * 3 = 1/3
1/1 * 3 + 1/3 * 5 = 2/5
1/1 * 3 + 1/3 * 5 + 1/5 * 7 = 3/7
1/1 * 3 + 1/3 * 5 + 1/5 * 7 + 1/7 * 9 = 4/9
From the given pattern, it can be observed that the numerator of each fraction in the sequence is simply increasing by 1 (starting from 1) with each subsequent term. The denominators follow an alternating pattern of increasing numbers, starting from 3 and continuing with odd numbers (3, 5, 7, 9, ...).
Based on this observation, we can express the general formula for the nth term as follows:
s(n) = (n / ((n * 2) - 1)) / ((n * 2) - 3)
Now, we need to prove this formula using mathematical induction:
Step 1: Base Case (n = 1)
When n = 1, the formula becomes:
s(1) = (1 / ((1 * 2) - 1)) / ((1 * 2) - 3)
= (1 / (2 - 1)) / (2 - 3)
= (1 / 1) / (-1)
= 1 / -1
= -1
The base case holds true.
Step 2: Inductive Hypothesis
Assume that the formula for s(k) is valid for some arbitrary positive integer k:
s(k) = (k / ((k * 2) - 1)) / ((k * 2) - 3)
Step 3: Inductive Step
We need to prove that the formula holds for n = k + 1:
s(k + 1) = ((k + 1) / (((k + 1) * 2) - 1)) / (((k + 1) * 2) - 3)
Expanding the equation:
s(k + 1) = ((k + 1) / (2k + 2 - 1)) / (2k + 2 - 3)
= ((k + 1) / (2k + 1)) / (2k - 1)
Now, we need to simplify the equation further:
s(k + 1) = ((k + 1) / (2k + 1)) / (2k - 1)
= ((k + 1) / (2k + 1)) * (1 / (2k - 1))
= (k + 1) / ((2k + 1) * (2k - 1))
= (k + 1) / (4k² - 1)
Since s(k) = (k / ((k * 2) - 1)) / ((k * 2) - 3) holds true, we can substitute it into the equation:
s(k + 1) = (k + 1) / (4k² - 1)
= (s(k) * (2k - 3)) / (2k - 1)
Therefore, the formula s(n) = (n / ((n * 2) - 1)) / ((n * 2) - 3) is valid for all positive integers n by mathematical induction.