Geometric Series

**classicstrings** · April 8th 2006, 06:03 AM

I have to find the nth term of this geometric series, i know the formula but im not sure if i need to simplify my answer.

Formula = tn = ar^(n-1)

a) 4,8,16,32 - this I can simplify and get 2^2 * 2^(n-1) which is 2^(n-1)
b) 6,24,96 - can you simplify this after you get 6 * 4^(n-1)?
c) 128,64,32, - same as above
d) 3,3root3, 9, 9root3... - SAA
e) 8,-4,2,-1 ... SAA
Thanks!

**Aradesh** · April 8th 2006, 06:28 AM

a) you say its $2^{n-1}$ but $2^2 \cdot 2^{n-1} = 2^{n+1}$ but thats probably what you meant.
b) $6 \cdot 4^{n-1}$ looks good. can't really simplify it because 4 doesn't divide 6.
c) 128,64,32... this is $2^{8-n}$
d) $3^{\frac{n+1}{2}}$
e) $-(-2)^{4-n}$ looks good.

**classicstrings** · April 9th 2006, 06:45 AM

Thanks for the answers but can you explain how to get them, especially c and d, as i dont know how to convert them (simplify them) when they get difficult. Thanks!

**classicstrings** · April 9th 2006, 11:26 PM

Still dont understand how you come to answers?!

**Aradesh** · April 10th 2006, 02:36 AM

well i presume you can do a and b.
i'll try talking through 'c' a little bit.
looking at:
128, 64, 32
i can immediately see that each number is being divided by two. infact i know them all to be powers of two. but you can check this by dividing one number by the next
128/64 = 2
64/32 = 2
dividing by two is the same as multiplying by a half so i could have expressed it as a common ratio as $\frac{1}{2}$ , however i thought it would be more tidy to use 'two'.
however with each step being divided by two instead of being multiplied, you have to take one from the index each step rather than adding one.
so i noticed that 128 = 2^7 then 64 = 2^6 and 32 = 2^5. notice that the indices are decreasing.
if you were using a common ratio of 1/2 you'd need to see that $128 = ( \frac{1}{2})^{-7}$ and $64 = (\frac{1}{2})^{-6}$ with increasing indices.
since the index decreases with each step i needed to use a negative n value. so i knew it'd be something like $2^{-n}$ . but if we just use this we see it doesn't add up..... since when n=1 the value is 128, or 2^7. so i had to adjust my result to get something so that this would work.
so when n is 1 with $2^{-n}$ i'd have $2^{-1}$ but i want $2^{7}$ so i had to add 8 to it, giving $2^{8-n}$ .
but if we had used the common ratio $\frac{1}{2}$ . we would have come up with $(\frac{1}{2})^{n-8}$

and we can see that this works for each one of the three values we were given:
n = 1, $2^{8-1} \equiv 2^7 \equiv 128$
n = 2, $2^{8-2} \equiv 2^6 \equiv 64$
n = 3, $2^{8-3} \equiv 2^5 \equiv 32$ .

now for d.
we have:
$3, \ 3 \sqrt{3}, \ 9$
here we can see that each step is being multiplied by $3^{\frac{1}{2}}$ . i don't know if this is difficult to see for you, maybe if you're not too comfortable with fractional indices.
we can see that the ratio from $3$ to $3 \sqrt{3}$ is $\sqrt{3}$ quite obviously.
when you have a square number you are multiplying two numbers together like 8*8 = 8^2. or if we squared 3^4.. we'd have $3^4 \times 3^4 \equiv (3^4)^2 \equiv 3^8$ , which is because when you multiply numbers you add the indices. so if you square a number in the form $n^k$ it becomes $n^k \cdot n^k = n^{2k}$ .
you can see that to get from a square number back to the original number you halve the index. so the square root of $n^{2k}$ is $n^k$ and the square root of $3^8$ is $3^4$ . likewise the square root of $3$ or $3^1$ is $3^{\frac{1}{2}}$ .
but anyway back to the problem.
we see that the common ratio is $\sqrt{3}$ which is the same as $3^{\frac{1}{2}}$ .
so i presume that the result is going to be something like
$(3^{\frac{1}{2}})^n = 3^{\frac{n}{2}}$ .
when n=1 this gives $3^{\frac{1}{2}} = \sqrt{3}$ which isn't what we want, we want 3 when n=1. so the index needs to be adjusted to be 1 when n=1. so we have
$3^{\frac{n}{2} + \frac{1}{2}} = 3^{\frac{n+1}{2}}$ .
which we can see to be true for the given values
n = 1, $3^{\frac{2}{2}} = 3$
n = 2, $3^{\frac{3}{2}} = 3\sqrt{3}$
n = 3, $3^{\frac{4}{2}} = 9$ .

e works pretty much the same however its utilizing the fact that a negative number to the power of an odd number ( $(-1)^{odd} = -1$ ) is odd, and that to the power of an even number, its even ( $(-1)^{even} = 1$ ).

**Aradesh** · April 10th 2006, 02:53 AM

actually, though. adjusting the indices as i have only has the same effect of multiplying each term of the ratio by an extra number.
if we want /everything/ in the form ar^(n-1) we will have to adjust the answers a little.
a) 4, 8, 16, 32
here using $2^{n-1}$ gives $2^0$ for n=1, so we want to multiply each value in the series by '4'. giving $4 \times 2^{n-1}$
which is the same as $2^{n+1}$ which i feel maybe isn't in the form of the result the problem asked for.
b) 6, 24, 96
we can see common ratio is 4.
with the starting value of '6' we get $6 \times 4^{n-1}$
c) 128, 64, 32
with a common ratio of $\frac{1}{2}$ and a starting value of 128 we get $128 \times (\frac{1}{2})^{n-1}$
d) 3, 3sqrt3, 9
comon ratio $\sqrt{3}$ and starting value of 3. giving us $3 \times (\sqrt{3})^{n-1}$
and finally e)
8,-4,2,-1...
the common ratio is -2
giving: $8 \times (-2)^{n-1}$

this seemed a fair bit easier didn't it! sorry i didn't notice this at first, however i still like my original answers

.
the reason an index of n-1 is used is because when n=1 (the first term) this makes the index zero, and anything to the power of zero is 1. so this means we just multiply the first term by the common ratio to the power of n-1.

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