Hello its my birthday please help me !!
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Answer:
Proofs contained within the explanation.
Step-by-step explanation:
These induction proofs will consist of a base case, assumption of the equation holding for a certain unknown natural number, and then proving it is true for the next natural number.
a)
Proof
Base case:
We want to shown the given equation is true for n=1:
The first term on left is 2 so when n=1 the sum of the left is 2.
Now what do we get for the right when n=1:
So the equation holds for n=1 since this leads to the true equation 2=2:
We are going to assume the following equation holds for some integer k greater than or equal to 1:
Given this assumption we want to show the following:
Let's start with the left hand side:
The first k terms we know have a sum of .5k(3k+1) by our assumption.
Distribute for the second term:
Combine terms in second term:
Factor out a half from both terms:
Distribute for both first and second term in the [ ].
Combine like terms in the [ ].
The thing inside the [ ] is called a quadratic expression. It has a coefficient of 3 so we need to find two numbers that multiply to be ac (3*4) and add up to be b (7).
Those numbers would be 3 and 4 since
3(4)=12 and 3+4=7.
So we are going to factor by grouping now after substituting 7k for 3k+4k:
.
Therefore for all integers n equal or greater than 1 the following equation holds:
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b)
Proof:
Base case: When n=1, the left hand side is 1.
The right hand at n=1 gives us:
So both sides are 1 for n=1, therefore the equation holds for the base case, n=1.
We want to assume the following equation holds for some natural k:
.
We are going to use this assumption to show the following:
Let's start with the left side:
We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:
Factor out the 1/4 from both of the two terms:
Combine the like terms inside the [ ]:
Apply law of exponents:
Therefore the following equation holds for all natural n:
.
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