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Norma-Jean [14]
4 years ago
13

What is the fourth term of the sequence a1=m an= 2an-1

Mathematics
1 answer:
kompoz [17]4 years ago
6 0

Answer:

  8m

Step-by-step explanation:

We assume you're describing a geometric sequence with first term m and common ratio 2. Then the n-th term is ...

  a[n] = m·2^(n-1)

and the fourth term is

  a[4] = m·2^(4-1) = 8m

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xeze [42]
<h3>Answer: 1</h3>

where x is nonzero

=======================================================

Explanation:

We'll use two rules here

  • (a^b)^c = a^(b*c) ... multiply exponents
  • a^b*a^c = a^(b+c) ... add exponents

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

The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.

Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)

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

After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2),  (b^2-c^2) and (c^2-a^2)

Add up those exponents (using rule 2 above) and we get

(a^2-b^2)+(b^2-c^2)+(c^2-a^2)

a^2-b^2+b^2-c^2+c^2-a^2

(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)

0a^2 + 0b^2 + 0c^2

0+0+0

0

All three exponents add to 0. As long as x is nonzero, then x^0 = 1

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