Question

I need help with these. They are hard.​

Answer 1

Answer:

Find the explicit from for the sequence $t_n=t_{n-1}+4,t=6$:

$a_n=4n+2$

This next question I edited a bit.  Your question just says find the four terms.  I'm assuming they meant the first four. I also changed the c to an $a$.

Find the first four terms of the sequence given by: $a_n=n a_{n-1}-3,a_1=2$:

a) 2,1,0.-3

You might want to read that second question again because there is errors in the question or things that don't really make sense.  I made my own interpretation of the problem based on my own mathematical experience.

Step-by-step explanation:

So your first question actually says that you can find a term by taking that term's previous term and adding 4.

So more terms of the sequence starting at first term 6 is:

6,10,14,18,....

This is an arithmetic sequence.  When thinking of arithmetic sequences you should just really by thinking about equations of lines.

Let's say we have this table for (x,y):

x  |   y

----------

1      6

2    10

3     14

4     18

So we already know the slope which is the common difference of an arithmetic sequence.

We also know point slope form of a line is $y-y_1=m(x-x_1)$ where m is the slope and $(x_1,y_1)$ is a point on the line.  You can use any point on the line. I'm going to use the first point (1,6) with my slope=4.

$y-6=4(x-1)$

$y=6+4(x-1)$    :I added 6 on both sides here.

$y=6+4x-4$     :I distribute here.

$y=4x+2$        :This is what I get after combining like terms.

So $a_n=y$ and $x=n$ so you have:

$a_n=4n+2$

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The first four terms of this sequence will be given by:

$a_1,a_2,a_3,a_4$

$a_1=2$ so it is between choice a, c, and d.

$a_n=na_{n-1}-3$

To find $a_2$ replace n with 2:

$a_2=2a_{1}-3$

$a_2=2(2)-3$

$a_2=4-3$

$a_2=1$

So we have to go another further the only one that has first two terms 2,1 is choice a.