Use mathematical induction to prove that the following is true.
8+11+14...+(3n+5)=1/2n(3n+13), for all n in the set of natural numbers.
prove that P(1) is true:
8 = 1/2 1(3*1+13) = 16/2 = 8
Assuming P(k), see what P(k+1) means:
8+11+...+(3k+5)+(3(k+1)+5) = k/2 (3k+13) + (3(k+1)+5)
= k/2 (3k+13) + 3k+8
1/2 (3k^2+13k + 6k+16)
= 1/2 (3k^2+19k+16)
= 1/2 (k+1)(3k+16)
= 1/2 (k+1)(3(k+1)+13)
= P(k+1)
So, P(1) and P(k) ==> P(k+1)
To prove that the equation 8+11+14+...+(3n+5)=(1/2)n(3n+13) is true for all n in the set of natural numbers using mathematical induction, we will follow these steps:
Step 1: Base Case
We'll start by checking if the equation holds true for the smallest possible value of n. In this case, let's take n=1.
Left-hand side (LHS):
For n = 1, the left-hand side (LHS) of our equation becomes:
8 = 1/2 * 1 * (3*1 + 13)
8 = 1/2 * (16)
8 = 8
The left-hand side (LHS) equals the right-hand side (RHS), so the equation is true for n = 1.
Step 2: Inductive Hypothesis
Assume that the equation is true for some arbitrary positive integer k. This is called the inductive hypothesis and we assume that it holds true for n = k.
So, our assumption is:
8 + 11 + 14 + ... + (3k + 5) = (1/2)k(3k + 13)
Step 3: Inductive Step
We need to prove that if the equation holds true for n = k, it also holds true for n = k + 1. This is called the inductive step.
For n = k+1, the left-hand side (LHS) becomes:
8 + 11 + 14 + ... + (3k + 5) + (3(k+1) + 5)
By the inductive hypothesis, we replace the sum up to 3k + 5 with (1/2)k(3k + 13):
(1/2)k(3k + 13) + (3(k+1) + 5)
Now, simplify the equation:
(1/2)k(3k + 13) + (3k + 3 + 5)
(1/2)k(3k + 13) + (3k + 8)
Now, factor out 3k+8:
(1/2)k(3k + 13 + 2) + (3k + 8)
(1/2)k(3k + 15) + (3k + 8)
Next, simplify the equation further:
(1/2)k(3k + 15) + 3k + 8
(1/2)k(3(k + 5)) + 3k + 8
(3k/2)(k+5) + 3k + 8
Now, combine the terms:
3k^2/2 + 15k/2 + 3k + 8
(3k^2 + 15k + 6k + 16)/2
(3k^2 + 21k + 16)/2
(3k^2+ 21k + 13k + 16)/2
(3k(k+7) + 13(k + 1))/2
Now, we see that we have a similar form on the right-hand side (RHS):
(1/2)(k+1)(3(k+1) + 13)
Therefore, the equation holds true for n = k+1 as well.
Step 4: Conclusion
We have shown that the equation is true for n = 1 (base case) and if it holds true for n = k, then it also holds true for n = k + 1 (inductive step). Therefore, by mathematical induction, the equation 8+11+14+...+(3n+5)=(1/2)n(3n+13) is true for all n in the set of natural numbers.
To prove the given statement using mathematical induction, we need to follow two steps:
Step 1: Base Case
Step 2: Inductive Step
Step 1: Base Case
We first need to check if the statement holds for the smallest value of n, which is usually n = 1.
Let's substitute n = 1 into the equation and simplify it:
8 + 11 + 14 + (3(1) + 5) = 1/2(1)(3(1) + 13)
8 + 11 + 14 + 8 = 1/2(3 + 13)
41 = 1/2(16)
41 = 8
Since 41 is not equal to 8, the base case is false, and the statement does not hold true for n = 1.
Therefore, we cannot proceed with the inductive step since the statement fails at the base case.