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Writing the terms of an alternating sequence defined by an explicit formula

Write the first five terms of the sequence.

a n = ( 1 ) n n 2 n + 1

Substitute n = 1 , n = 2 , and so on in the formula.

n = 1 a 1 = ( 1 ) 1 2 2 1 + 1 = 1 2 n = 2 a 2 = ( 1 ) 2 2 2 2 + 1 = 4 3 n = 3 a 3 = ( 1 ) 3 3 2 3 + 1 = 9 4 n = 4 a 4 = ( 1 ) 4 4 2 4 + 1 = 16 5 n = 5 a 5 = ( 1 ) 5 5 2 5 + 1 = 25 6

The first five terms are { 1 2 , 4 3 ,− 9 4 , 16 5 ,− 25 6 } .

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In [link] , does the (–1) to the power of n account for the oscillations of signs?

Yes, the power might be n , n + 1 , n 1 , and so on, but any odd powers will result in a negative term, and any even power will result in a positive term.

Write the first five terms of the sequence:

a n = 4 n ( 2 ) n

The first five terms are { 2 ,   2 ,   3 2 ,   1 ,   5 8 } .

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Investigating piecewise explicit formulas

We’ve learned that sequences are functions whose domain is over the positive integers. This is true for other types of functions, including some piecewise functions . Recall that a piecewise function is a function defined by multiple subsections. A different formula might represent each individual subsection.

Given an explicit formula for a piecewise function, write the first n terms of a sequence

  1. Identify the formula to which n = 1 applies.
  2. To find the first term, a 1 , use n = 1 in the appropriate formula.
  3. Identify the formula to which n = 2 applies.
  4. To find the second term, a 2 , use n = 2 in the appropriate formula.
  5. Continue in the same manner until you have identified all n terms.

Writing the terms of a sequence defined by a piecewise explicit formula

Write the first six terms of the sequence.

a n = { n 2 if  n  is not divisible by 3 n 3 if  n  is divisible by 3

Substitute n = 1 , n = 2 , and so on in the appropriate formula. Use n 2 when n is not a multiple of 3. Use n 3 when n is a multiple of 3.

a 1 = 1 2 = 1 1 is not a multiple of 3 .  Use  n 2 . a 2 = 2 2 = 4 2 is not a multiple of 3 .  Use  n 2 . a 3 = 3 3 = 1 3 is a multiple of 3 .  Use  n 3 . a 4 = 4 2 = 16 4 is not a multiple of 3 .  Use  n 2 . a 5 = 5 2 = 25 5 is not a multiple of 3 .  Use  n 2 . a 6 = 6 3 = 2 6 is a multiple of 3 .  Use  n 3 .

The first six terms are { 1 ,   4 ,   1 ,   16 ,   25 ,   2 } .

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Write the first six terms of the sequence.

a n = { 2 n 3 if  n  is odd 5 n 2 if  n  is even

The first six terms are { 2 ,   5 ,   54 ,   10 ,   250 ,   15 } .

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Finding an explicit formula

Thus far, we have been given the explicit formula and asked to find a number of terms of the sequence. Sometimes, the explicit formula for the n th term of a sequence is not given. Instead, we are given several terms from the sequence. When this happens, we can work in reverse to find an explicit formula from the first few terms of a sequence. The key to finding an explicit formula is to look for a pattern in the terms. Keep in mind that the pattern may involve alternating terms, formulas for numerators, formulas for denominators, exponents, or bases.

Given the first few terms of a sequence, find an explicit formula for the sequence.

  1. Look for a pattern among the terms.
  2. If the terms are fractions, look for a separate pattern among the numerators and denominators.
  3. Look for a pattern among the signs of the terms.
  4. Write a formula for a n in terms of n . Test your formula for n = 1 ,   n = 2 , and n = 3.

Writing an explicit formula for the n Th term of a sequence

Write an explicit formula for the n th term of each sequence.

  1. { 2 11 , 3 13 , 4 15 , 5 17 , 6 19 , }
  2. { 2 25 , 2 125 , 2 625 , 2 3 , 125 , 2 15 , 625 , }
  3. { e 4 , e 5 , e 6 , e 7 , e 8 , }

Look for the pattern in each sequence.

  1. The terms alternate between positive and negative. We can use ( 1 ) n to make the terms alternate. The numerator can be represented by n + 1. The denominator can be represented by 2 n + 9.

    a n = ( 1 ) n ( n + 1 ) 2 n + 9

  2. The terms are all negative.

    So we know that the fraction is negative, the numerator is 2, and the denominator can be represented by 5 n + 1 .

    a n = 2 5 n + 1
  3. The terms are powers of e . For n = 1 , the first term is e 4 so the exponent must be n + 3.

    a n = e n + 3
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Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6
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