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2 years ago

Determine if the sequence below is arithmetic or geometric and determine the common difference / ratio in simplest form.

Mathematics

1 answer:

andreev551 [17]2 years ago

5 0

Answer:

Step-by-step explanation:

This is an arithmetic sequence and the common difference = - 2

You might be a little puzzled by the minus sign in front of the two. You can look at the sequence for the answer. It is going down. A plus 2 would make it go up. Minus signs are always a bit tricky and you have to watch out what you do with them. But the rule is simple

a minus common difference produces a sequence that goes down.
a plus common difference produces a squence that goes up.

You know that this is an arithmetic sequence because the common difference can be found by adding or subtracting a number from the previous term to get to the next term.

A geometrice sequence would use multiplication or division to get from one term to the next.

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Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Factor this expression.<br> x2 + 9x + 8

Ket [755]

Answer:

(x+1)(x+8) When factoring squares whose squared coefficient is one the roots must add up to the coefficient of the slope and multiply out to the intercept value.

6 0

3 years ago

Please please help me out!!!!!!!!!!!!!!!!!!!!!!!!!!

nexus9112 [7]

I hope it can help you