MHB How Do You Solve a Geometric Sum with Alternating Signs?

AI Thread Summary
To solve the geometric sum with alternating signs, the sum S is defined as S = 3 - 3/2 + 3/4 - 3/8 + 3/16 - 3/32 + ... - 3/128. The common ratio is identified as r = -1/2, and the first term is a = 3. The sum can be calculated using the geometric series formula, which requires determining the number of terms, n, and the common ratio. The series can be expressed as S = 3 * Σ from j=0 to 7 of (-1/2)^j. The solution involves applying the geometric sum formula to find the total value of S.
Kola Citron
Messages
1
Reaction score
0
Hey!

I'm stuck again and not sure how to solve this question been at it for a few hours. Any help is appreciated as always.

Q: (1) Let the sum S = 3- 3/2 + 3/4 - 3/8 + 3/16 - 3/32 +...- 3/128. Determine integers a , n and a rational number k so that...(Image)

r/askmath - Summation and geometric sums

(2 )And then calculate S using the geometric sum formula.

Thank you!
 
Mathematics news on Phys.org
common ratio is $r = -\dfrac{1}{2}$

note $128 = 2^7$

first term is $a = 3$

$\displaystyle S = 3 \sum_{j=0}^7 \left(-\dfrac{1}{2}\right)^j$

you can calculate the sum ...
 
Given that it is a "geometric sum", a+ ar+ ar^2+ ...,, you can determine r, the "common ratio" by just dividing the second term by the first: ar/a= r. In this problem that is (-3/2)/3= -1/2.
 
Thread 'Video on imaginary numbers and some queries'
Hi, I was watching the following video. I found some points confusing. Could you please help me to understand the gaps? Thanks, in advance! Question 1: Around 4:22, the video says the following. So for those mathematicians, negative numbers didn't exist. You could subtract, that is find the difference between two positive quantities, but you couldn't have a negative answer or negative coefficients. Mathematicians were so averse to negative numbers that there was no single quadratic...
Thread 'Unit Circle Double Angle Derivations'
Here I made a terrible mistake of assuming this to be an equilateral triangle and set 2sinx=1 => x=pi/6. Although this did derive the double angle formulas it also led into a terrible mess trying to find all the combinations of sides. I must have been tired and just assumed 6x=180 and 2sinx=1. By that time, I was so mindset that I nearly scolded a person for even saying 90-x. I wonder if this is a case of biased observation that seeks to dis credit me like Jesus of Nazareth since in reality...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Back
Top