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muminat
2 years ago
13

Mathematical induction, prove the following two statements are true

Mathematics
1 answer:
adelina 88 [10]2 years ago
3 0
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
--------------------------------------------------------------------------

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}}
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Give the first ten terms of the following sequences. You can assume that the sequences start with an index of 1. Logs are to bas
Vitek1552 [10]

Answer:

a)

a1 = log(1) = 0 (2⁰ = 1)

a2 = log(2) = 1 (2¹ = 2)

a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849

a4 = log(4) = 2 (2² = 4)

a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322

a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)

a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807

a8 = log(8) = 3 (2³ = 8)

a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )

a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322

b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:

log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that

a1 = 1

a2 = 2

a3 = 3

a4 = 4

a5 = 5

a6 = 6

a7 = 7

a8 = 8

a9 = 9

a10 = 10

I hope this works for you!!

3 0
2 years ago
Problem of the Week
Ierofanga [76]

Answer:

CD is 6 cm and DE is 4 cm.

Step-by-step explanation:

Well it‘s the correct answer and I got it right so good luck out there bois!

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2 years ago
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We have to find the values of m and n. We know that the value of n is 5 times as much as the value of m : n = 5 m. Also n is 36 more than the value of m : n = m + 36. This is a system of equations. Of we substitute n = 5 m into the second equation: 5 m = m + 36; 5 m - m = 36; 4 m = 36; m = 36 : 4; m = 9. Finally: n = 5 * 9 ; n = 45. Answer: The values are n = 45 and m = 9.
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