Mathematics, 18.10.2019 00:50 cfonse11

Mathematical induction, prove the following two statements are true

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Answer from: Clerry

Prove:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}$
____________________________________________

Base Step: For n=1:
$n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1$
and
$4-\dfrac{n+2}{2^{n-1}}=4-3=1$
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Induction Hypothesis: Assume true for n=k. Meaning:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}$
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
$1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}$

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
$=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}$

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
$1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}$

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
$=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}$

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
$=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}$

Distribute the -2 and combine the fractions together,
$=4+\dfrac{-2k-4+(k+1)}{2^{k}}$

Combine like-terms,
$=4+\dfrac{-k-3}{2^{k}}$

pull the negative back out,
$=4-\dfrac{k+3}{2^{k}}$

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
$=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}$

Answer from: Quest

step-by-step explanation:

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

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