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11Alexandr11 [23.1K]
1 year ago
6

Find a formula for the general term an of the sequence, assuming that the pattern of the first few terms continues. (Assume that

n begins with 1.) −5, 10 3 , − 20 9 , 40 27 , − 80 81 , ...
Mathematics
1 answer:
mezya [45]1 year ago
8 0

Answer:

The formula to the sequence

-5, 10/3, -20/9, 40/27, -80/81, ...

is

(-1)^n. 5×2^(n-1). 3^(1-n)

For n = 1, 2, 3, ...

Step-by-step explanation:

The sequence is

-5, 10/3, -20/9, 40/27, -80/81, ...

By inspection, we see the following

- The numbers are alternating between - and +

- The numerator of a number is twice the numerator of the preceding number. The first number is 5.

- The denominator of a number is 3 raised to the power of (1 minus the position of the number)

Using these, we can write a formula for the sequence.

(-1)^n for n = 1, 2, 3, ... takes care of the alternation between + and -

5×2^(n-1) for n = 1, 2, 3, ... takes care of the numerators 5, 10, 20, 40, ...

3^(1-n) for n = 1, 2, 3, ... takes care of the denominators 1, 3, 9, 27, ...

Combining these, we have the formula to be

(-1)^n. 5×2^(n-1). 3^(1-n)

For n = 1, 2, 3, ...

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The answer is 1/4 because 7/12 is larger than 1/2 but 1/4 is 2x smaller

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2 years ago
What is 89,659 rounded to the nearest hundred thousand?
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<u><em>Answer:</em></u>

100,000


<u><em>Explanation:</em></u>

To round a number to the nearest hundred thousands, we need to check the digit in the ten thousands position:

1- If this digit is <u>less than 5</u>, we will round down. This means that the digit in the hundred thousands position will remain the same and all digits after it will be converted to zeroes

2- If this digit is <u>equal to or greater than 5</u>, we will round up. This means that we will add one to the digit in the hundred thousands position and convert all digit after it to zeroes


Now, the given number is:

89,659

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The digit in the ten thousands position is 8 which is greater than 5. Therefore, we will round up following rule 2 written above

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Hope this helps :)

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Which of the following is the expansion of (3c + d2)6?
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Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

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⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

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⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

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