Finding sum of a alternating geometric series

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Vikash

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The question is...for example if you consider the geometric series (-5/2 +5/4-5/8....5/32768) and asked to find the sum to 3 d.p but when I use the formula
a Un=ar(n-1) since there is a negative ratio I get a negative logarithm to solve which is impossible...is there another approach I can take. Please help..Your help will be worthwhile....
 
You can factor the 5 out of the computation. Now try to write the sum as an alternative geometric series and then I think you will know what to do.
 
[MATH]S = \dfrac{a(1-r^n)}{1-r}[/MATH]
where [MATH]a[/MATH] is the first term of the series and [MATH]r[/MATH] is the common ratio

fyi, [MATH]2^{15} = 32768[/MATH]
 
Vikash said:
The question is...for example if you consider the geometric series (-5/2 +5/4-5/8....5/32768) and asked to find the sum to 3 d.p but when I use the formula
a Un=ar(n-1) since there is a negative ratio I get a negative logarithm to solve which is impossible...is there another approach I can take. Please help..Your help will be worthwhile....
Click to expand...
Why are you taking logarithm of this beast?

Do you see that:

a = -5/2 &

r = - 1/2
 
Vikash said:
The question is...for example if you consider the geometric series (-5/2 +5/4-5/8....5/32768) and asked to find the sum to 3 d.p but when I use the formula
a Un=ar(n-1) since there is a negative ratio I get a negative logarithm to solve which is impossible...is there another approach I can take. Please help..Your help will be worthwhile....
Click to expand...
It looks like you're trying to solve the formula for the nth term to find n. (You don't mean "negative logarithm", but "logarithm of a negative number".) You can overcome this by ignoring signs.

You've been given both a hint for doing that without logs, and the formula for the sum of the finite series, in post #3.
 
Dr.Peterson said:
It looks like you're trying to solve the formula for the nth term to find n. (You don't mean "negative logarithm", but "logarithm of a negative number".) You can overcome this by ignoring signs.

You've been given both a hint for doing that without logs, and the formula for the sum of the finite series, in post #3.
Click to expand...
Thanks?
 
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