Studying for Edexcel January C2. Question on geometric series sum..?

Question

The sum of a series of the first n terms of a series is
(4n +5)2 {The two indicates squared}. Find the nth term of the series....

I though that Un= Sn - S(n-1)... But its not working ... Help!

sahsjing, please explain method you used!!!!!!!! If you do, I'll give you best answer! Thanks.

Actually sahsjing, sorry, I take offer back. Your method produces same series I got, which as someone else indicated, is neither geometric or arithmaic. Must be the whole question ia wrong. Teacher was a big mo**er fu**er any way. He was always drunk. God I hated him. Thanks any way guys.

Answer 1

Sum=(4n+5)^2

if n=1
the first term is 81

if n=2
the second term is 88

13^2-81=169-81= 88

if n=3
te third term is 120

That is, the series
81,88,120,...
is neither geometric nor arithmetic.

Check your question

Answer 2

A geometric series is formed by multiplying each term by a constant.
For example
each term is doubled
This series can be broken down to
So that each term can be written as
With a as the starting term and r as the multiplying factor.
r is called the common ratio.



A second example.
Write the first 4 terms of the geometric series described by

Substitute into




Finding the Sum of a Geometric Series

The following, ingenious method is used to find the sum of a geometric seriesWrite out Sn
Multiply by r. Notice how most terms pair up.
Subtract the two equations.
Factorize the left hand side.
Divide by r-1

Either arrangement will work, but the second is more appropriate if r<1





Example: Find the sum of the first 6 terms of the series
Define a, r and n
Substitute into formula.


The Sum To Infinity
In the case of
then

Our sum formula then becomes
For example: Find the sum of the first 8 terms of the following series and the sum to infinity
Define a, r, n
Substitute into formula

For the sum to infinity

Note that
so the series was not far from the sum to infinity after 8 terms.

Answer 3

First pair up the words: a million - a million/2 = a million/2 a million/4 - a million/8 = a million/8 a million/sixteen - a million/32 = a million/32 So your sequence is such as: a million/2 + a million/8 + a million/32 ... The nth term interior the sequence then is: a million/(2^(2n - a million)) Now check out the sequence so a tactics: S = a million/2 + a million/8 + a million/32 + .... a million/(2^(2n - a million)) Multiply this via a million/4: S * (a million/4) = a million/8 + a million/32 + ...... + a million/(2^(2n - a million)) + a million/(2^(2(n+a million) -a million) Is you subtract one form the different, all words however the 1st in one expression and the final in the different cancel: S - S*(a million/4) = a million/2 - a million/(2^(2(n+a million) -a million) S*(3/4) = a million/2 - a million/(2^(2(n+a million) -a million) S = (4/3)*(a million/2 - a million/(2^(2(n+a million) -a million)) that's right for any n. As n gets super, the final words has a tendency to 0 so for massive n: S = (4/3)*(a million/2) = 2/3 try some words to be sure it extremely sums to this: a million - a million/2 + a million/4 - a million/8 + a million/sixteen - a million/32 = 21/32 = 0.656 seems sensible

Answer 4

You are right.

The nth term = sum of the first n terms - sum of the first (n-1) terms: Un = Sn - S(n-1)

Un = Sn - S(n-1)
= (4n+5)^2 - (4(n-1)+5)^2
= (8n+6)(4), (used a^2 - b^2 = (a+b)(a-b))
= 8(4n+3)

Answer 5

For the following...

a = first term
r = common ratio
n= term no


n th term = ar^(n-1)


Sum of n terms

Sn = [a(1-r^n)]/(1-r)

for a convergent series

Sn = a/(1-r) where |r| < 1

Answer 6

try this
expand sum = (4n+5)^2
= 16n^2+40n+25 differentiate
nth term = 32n+40

just a thought.

Answer 7

sub in 1 to get the first term. Sub in 2 to get the first+second term, subtract to get the second term by itself. Do this a bit to get a few terms, find the pattern, then write it in the form

Nth = a + (n-1)*d

At least I think that's how you need to do it, try it and see.

Answer 8

what you need is a Graphic Calculator, the ones' that are banned from the exam hall!!!