GRE Quantitative Reasoning - Sequences

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

$\small R=4$

$\small \small R=\frac{1}{4}$

$\small \small R=\infty$

$\small \small R=2$

$\small R=0$

Explanation

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

$\large \large \large 2.929$

$\large \large 2.9289$

$\large \large 2.92903$

$\large 2.92897$

Explanation

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

Find the radius of convergence for the power series

$\small \sum_{n=0}^{\infty}\frac{(n!)^2x^n}{(2n)!}$

$\small R=4$

$\small \small R=\frac{1}{4}$

$\small \small R=\infty$

$\small \small R=2$

$\small R=0$

Explanation

We can use the limit

$\small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|$

to find the radius of convergence. We have

$\small \small \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right|=\lim \left| \frac{x^{n+1}[(n+1)!]^2}{(2(n+1))!} \frac{(2n!)}{(n!)^2x^n}\right|$

$\small \small \small \small =\lim |x|\left| \frac{(2n)!}{(2n+2)!} \frac{[(n+1)!]^2}{(n!)^2}\right|=\lim |x|\left| \frac{(n+1)(n+1)}{(2n+1)(2n+2)}\right|=\frac{|x|}{4}<1$

$\small \implies |x|<4=R$

This means the radius of convergence is $\small R=4$ .

Evaluate: $\large \large \sum_{x=1}^{10} \frac {1}{n}$ . (Round to 4 places)

$\large \large \large 2.929$

$\large \large 2.9289$

$\large \large 2.92903$

$\large 2.92897$

Explanation

Step 1: Plug in values into the function and add up the fraction:

$\large 1+\frac {1}{2}+\frac {1}{3}+\frac {1}{4}+\frac {1}{5}+\frac {1}{6}+\frac {1}{7}+\frac {1}{8}+\frac {1}{9}+\frac {1}{10}$

Step 2: Find the sum of the fractions....

We can convert the fractions to decimals:

$\large 1+.5+.3333+.25+.2+.1666+.142857+.125+.1111+.1$

$\large =2.92896$

Step 3: Round to $\large 4$ places...

$\large \large 2.92896 \implies 2.9290\implies 2.929$

If the first term of an arithmetic sequence is 2 and the third term is 8, find the $\large \large 16$ th term.

$\large 47$

$\large 49$

$\large 46$

$\large 43$

Explanation

Step 1: Find the difference between each term...

Subtract the first term from the third term...

$\large 8-2=6$
There are two terms between first and third...Take the answer in step 1 and divide by 2 to get the difference between consecutive terms...

$\large \frac {6}{2}=3$

The common difference is $\large 3$ .

Step 2: Find an equation that describes the sequence....

The equation is $\large 3n+2$ , where $\large n$ represents how many terms I need to calculate and $\large 2$ is the first term...

Step 3: Plug in $\large n$ ...

To find n, we subtract the term that we want from the original term...

So, if we want the $\large 16$ th term and we are given the first term...

Then $\large n=16-1=15$

So, $\large 3(15)+2=45+2=47$

The $\large 16$ th term is $\large 47$ .

If a sequence is 9, 3, -3, -9, -15, ...

What is the 10th term of the sequence?

Explanation

It will be helpful if you can see that the sequence is changing by -6 to get to each number. The quickest way to solve this is to keep going with the sequence by subtracting six from each number to get to the next number.

9, 3, -3, -9, -15, -21, -27, -33, -39, -45

-45 is the 10th number in the sequence.

Another way to find the 10th term is by using the equation for arithmetic sequences.

where is the term in question, is the first term, is the numbered term, and is the difference of the terms.

To find d subtract the first two terms.

Therefore, the equation becomes

$a_{10}=9+(10-1)(-6)$

$\a_{10}=9+(9)(-6)\a_{10}=9-54\a_{10}=-45$

Find the next term in the sequence:

Explanation

Step 1: Find the next term in the sequence:

Step 2: Can you recognize the sequence here??

This is the Fibonacci Sequence.. The sum of the previous two terms is equal to the next consecutive term..

The missing term is 13.

If the first terms of a sequence are ,find the th term.

$1251.\overline {11}$

Explanation

Step 1: Find the successive difference rows until we get equal values between every two consecutive numbers.

Step 2: Since we obtain equal values for the successive differences in the second row, hence the $n^{th}$ term of the sequence is a second-degree polynomial.

So, the $n^{th}$ term takes the form of

Step 3: Now, substituting into the formula, we get:

Step 4: Solve the system of linear equations:

So, the $n^{th}$ term takes the form:

Step 5: Plug in into the form...

The first three terms of a arithmetic sequence are . What is the term of this sequence?

Explanation

Step 1: We need to find the difference between the terms. To find the difference, subtract the first two terms.

Difference=

Step 2: To find the next term of a arithmetic sequence, we add the difference to the previous term.

Fourth term= term+5
term=. The fourth term is 19.

Step 3: To find the term, we must add a multiple of to the first term. In this problem, we are given the term, so we need to find how many terms are in between the and term.

. This tells me that I have to add fourteen times to get to the term. .

Step 4: Now that we know what the value of n, we can plug it in to the equation:

, where 4 is the starting number, and n represents how many times I need to add 5 to the next term (and up to the 15th term).

Let's plug it in:

So, the term in this sequence is .

If the first term of an arithmetic sequence is 2 and the third term is 8, find the $\large \large 16$ th term.

$\large 47$

$\large 49$

$\large 46$

$\large 43$

Explanation

Step 1: Find the difference between each term...

Subtract the first term from the third term...

$\large 8-2=6$
There are two terms between first and third...Take the answer in step 1 and divide by 2 to get the difference between consecutive terms...

$\large \frac {6}{2}=3$

The common difference is $\large 3$ .

Step 2: Find an equation that describes the sequence....

The equation is $\large 3n+2$ , where $\large n$ represents how many terms I need to calculate and $\large 2$ is the first term...

Step 3: Plug in $\large n$ ...

To find n, we subtract the term that we want from the original term...

So, if we want the $\large 16$ th term and we are given the first term...

Then $\large n=16-1=15$

So, $\large 3(15)+2=45+2=47$

The $\large 16$ th term is $\large 47$ .

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