Question

Please help solve these proofs asap!!!

Answer 1

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:

$\frac{1}{2}(1)(3(1)+1)$

$\frac{1}{2}(3+1)$

$\frac{1}{2}(4)$

$2$

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:

$2+5+8+\cdots+(3k-1)=\frac{1}{2}k(3k+1)$

Given this assumption we want to show the following:

$2+5+8+\cdots+(3(k+1)-1)=\frac{1}{2}(k+1)(3(k+1)+1)$

Let's start with the left hand side:

$2+5+8+\cdots+(3(k+1)-1)$

$2+5+8+\cdots+(3k-1)+(3(k+1)-1)$

The first k terms we know have a sum of .5k(3k+1) by our assumption.

$\frac{1}{2}k(3k+1)+(3(k+1)-1)$

Distribute for the second term:

$\frac{1}{2}k(3k+1)+(3k+3-1)$

Combine terms in second term:

$\frac{1}{2}k(3k+1)+(3k+2)$

Factor out a half from both terms:

$\frac{1}{2}[k(3k+1)+2(3k+2]$

Distribute for both first and second term in the [ ].

$\frac{1}{2}[3k^2+k+6k+4]$

Combine like terms in the [ ].

$\frac{1}{2}[3k^2+7k+4$

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:

$\frac{1}{2}[3k^2+3k+4k+4]$

$\frac{1}{2}[3k(k+1)+4(k+1)]$

$\frac{1}{2}[(k+1)(3k+4)]$

$\frac{1}{2}(k+1)(3k+4)$

$\frac{1}{2}(k+1)(3(k+1)+1)$.

Therefore for all integers n equal or greater than 1 the following equation holds:

$2+5+8+\cdots+(3n-1)=\frac{1}{2}n(3n+1)$

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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:

$\frac{1}{4}(5^1-1)$

$\frac{1}{4}(5-1)$

$\frac{1}{4}(4)$

$1$

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:

$1+5+5^2+\cdots+5^{k-1}=\frac{1}{4}(5^k-1)$.

We are going to use this assumption to show the following:

$1+5+5^2+\cdots+5^{(k+1)-1}=\frac{1}{4}(5^{k+1}-1)$

Let's start with the left side:

$1+5+5^2+\cdots+5^{(k+1)-1}$

$1+5+5^2+\cdots+5^{k-1}+5^{(k+1)-1}$

We know the sum of the first k terms is 1/4(5^k-1) given by our assumption:

$\frac{1}{4}(5^k-1)+5^{(k+1)-1}$

$\frac{1}{4}(5^k-1)+5^k$

Factor out the 1/4 from both of the two terms:

$\frac{1}{4}[(5^k-1)+4(5^k)]$

$\frac{1}{4}[5^k-1+4\cdot5^k]$

Combine the like terms inside the [ ]:

$\frac{1}{4}(5 \cdot 5^k-1)$

Apply law of exponents:

$\frac{1}{4}(5^{k+1}-1)$

Therefore the following equation holds for all natural n:

$1+5+5^2+\cdots+5^{n-1}=\frac{1}{4}(5^n-1)$.

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