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C1 (SeriesSequences) questions

Having a bit of trouble with these;

1) Find in a simplified form ∑r=1n(7r−2)\displaystyle\sum_{r=1}^n (7r-2)
What is this actually asking for?

2) Someone invests £20 one month, then 25, then 30, then 35 and so on until he has made an investment £500. How much has she invested in total?

Is this using
Number of terms x sum of first and last / 2
So (96 x 520)/2 = £24 960

3) Find the first 4 terms in the sequences whos nth term is given by
Un = nn+1\frac{n}{n+1}

4) Find the first 4 terms of sequence defined as follows;
tn+1=3tn−1[br]t1=4t_{n+1} = 3t_n -1[br]t_1 = 4

5)3 consecutive terms of an arithmetic sequence are

3p-2, 4p-6, 9p+16

Find the value of p and the common difference.

-----------

+rep to help, even if its just the formula to use etc.
Original post by jbeach09
Having a bit of trouble with these;

1) Find in a simplified form ∑r=1n(7r−2)\displaystyle\sum_{r=1}^n (7r-2)
What is this actually asking for?



I'm only going to look at the first one. :tongue:

It says what it wants but to help...

∑r=1n(7r−2)=7∑r=1nr−∑r=1n2\displaystyle\sum_{r=1}^n (7r-2) =7\displaystyle\sum_{r=1}^n r-\displaystyle\sum_{r=1}^n 2 and you should know what ∑r=1nr\displaystyle\sum_{r=1}^n r is.
Reply 2
Avatar for Expendable
Expendable
OP
11
Original post by Get me off the £\?%!^@ computer
I'm only going to look at the first one. :tongue:

It says what it wants but to help...

∑r=1n(7r−2)=7∑r=1nr−∑r=1n2\displaystyle\sum_{r=1}^n (7r-2) =7\displaystyle\sum_{r=1}^n r-\displaystyle\sum_{r=1}^n 2 and you should know what ∑r=1nr\displaystyle\sum_{r=1}^n r is.


I got that far with the summation results page in my book, but I've got no idea what the next step is.

I was absent through most of the series lessons due to illness and have had this assignment, the questions with find the nth term are ok because I was in that, but these I've got no idea.

Surely that's saying the last value of r in the sequence is n, and the first value is 1, but whats the method next?
Original post by jbeach09
I got that far with the summation results page in my book, but I've got no idea what the next step is.

I was absent through most of the series lessons due to illness and have had this assignment, the questions with find the nth term are ok because I was in that, but these I've got no idea.

Surely that's saying the last value of r in the sequence is n, and the first value is 1, but whats the method next?


∑r=1nr=n2(n+1)\displaystyle\sum_{r=1}^n r=\frac{n}{2}(n+1)

Post some working if you need any more help. :smile:
Reply 4
Avatar for Oasis Jnr
Oasis Jnr
i also hate this topic but for the second one,
you use the formula : Sn: n/2(a +l)
so ur first term is 20 and your last term is 500, sub in those and work your way to the anser

for the 3rd one, just sub in 1 for n and do the same for 2 and 3 and 4 as they want the first 4

for the 4th one, ur meant to be using the recurrance method, so sub in 1 for n and they tell u that t1=4 so 4 times 3 and then minus one. then u will get t2 and use t2 to find t3 and so forth

the fifth and the first one are tricky
good luck
Reply 5
Avatar for Expendable
Expendable
OP
11
Thanks, managed to do them all apart from the last one and first one now :smile:

Original post by Get me off the £\?%!^@ computer
∑r=1nr=n2(n+1)\displaystyle\sum_{r=1}^n r=\frac{n}{2}(n+1)

Post some working if you need any more help. :smile:

So would you do

7(n2(n+1))−n2(n+1)27(\frac{n}{2}(n+1))-\frac{n}{2}(n+1)2
or am I barking up the wrong tree?

If I still cannot get it, it's probably worth me going and having a word with my teacher tomorrow.
Original post by jbeach09
Thanks, managed to do them all apart from the last one and first one now :smile:


So would you do

7(n2(n+1))−n2(n+1)27(\frac{n}{2}(n+1))-\frac{n}{2}(n+1)2
or am I barking up the wrong tree?

If I still cannot get it, it's probably worth me going and having a word with my teacher tomorrow.


This bit

∑r=1n2\displaystyle\sum_{r=1}^n 2

is just nx2.

Yes talking to your teacher is a good idea. :smile:

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