md12110332
Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

2/3+(2/3)^2+(2/3)^3 + ... = x, find x.

Sagot :

LammettHash

Let S be the sum of the first n terms of the left side:

[tex]S = \dfrac23 + \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \cdots + \left(\dfrac23\right)^n[/tex]

Multiply both sides by 2/3 :

[tex]\dfrac23 S = \left(\dfrac23\right)^2 + \left(\dfrac23\right)^3 + \left(\dfrac23\right)^4 + \cdots + \left(\dfrac23\right)^{n+1}[/tex]

Subtract this from S :

[tex]S - \dfrac23 S = \dfrac23 - \left(\dfrac23\right)^{n+1}[/tex]

Solve for S :

[tex]\dfrac13 S = \dfrac23 - \left(\dfrac23\right)^{n+1}[/tex]

[tex]S = 2 - 3 \left(\dfrac23\right)^{n+1}[/tex]

As n gets larger and larger, S converges to the given sum, and the term (2/3)ⁿ⁺¹ converges to zero, which leaves us with

[tex]\displaystyle \lim_{n\to\infty} S = \boxed{x = 2}[/tex]

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.