BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

A = (1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8). What is the value of 1 - 2/

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

A = (1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8). What is the value of 1 - 2/

by Max@Math Revolution » Wed Apr 22, 2020 7:04 am
[GMAT math practice question]

A = (1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8). What is the value of 1 - 2/3A?

A. 1
B. 3
C. 1/3
D. 1/3^16
E. 1/3^20
Top
Source: — Problem Solving |
User avatar
Elite Legendary Member
Posts: 3991
Joined: Fri Jul 24, 2015 2:28 am
Location: Las Vegas, USA
Thanked: 19 times
Followed by:37 members

Re: A = (1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8). What is the value of 1 - 2/

by Max@Math Revolution » Fri Apr 24, 2020 1:31 am
=>

2/3A = (1 - 1/3)(1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8) = (1-1/3^2)(1+1/3^2)(1+1/3^4)(1+1/3^8)
= (1-1/3^4)(1+1/3^4)(1+1/3^8)
= (1-1/3^8)(1+1/3^8)
= 1-1/3^16

Thus, we have 1 - 2/3A = 1 – (1-1/3^16) = 1/3^16.

Therefore, D is the answer.
Answer: D
Top
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pm
Followed by:5 members

Re: A = (1 + 1/3)(1 + 1/3^2)(1 + 1/3^4)(1 + 1/3^8). What is the value of 1 - 2/

by deloitte247 » Sat Apr 25, 2020 1:50 pm
$$A=\left(1+\frac{1}{3}\right)\left(1+\frac{1}{3^2}\right)\left(1+\frac{1}{3^4}\right)\left(1+\frac{1}{3^8}\right)$$
$$A=\left(\frac{3+1}{3}\right)\left(\frac{3^2+1}{3^2}\right)\left(\frac{3^4+1}{3^4}\right)\left(\frac{3^8+1}{3^8}\right)$$
$$A=\left(\frac{3}{3}+\frac{1}{3}\right)\left(\frac{3^2}{3^2}+\frac{1}{3^2}\right)\left(\frac{3^4}{34}+\frac{1}{3^4}\right)\left(\frac{3^8}{3^8}+\frac{1}{3^8}\right)$$
$$A=\left(\frac{1}{3}\right)\left(\frac{1}{3^2}\right)\left(\frac{1}{3^4}\right)\left(\frac{1}{3^8}\right)$$
$$A=\left(\frac{1}{3^{1+2+4+8}}\right)\ =\frac{1}{3^{15}}$$
$$Therefore,\ 1-\frac{2}{3}A$$
$$Expres\sin g\ \frac{2}{3}\ in\ form\ of\ \ \frac{1}{3}$$
$$1-\left(1-\frac{1}{3}\right)A\ \ \left(because\ \frac{1}{1}-\frac{1}{3}=\frac{2}{3}\right)$$
$$Since\ A=\frac{1}{3^{15}}$$
$$=>\ 1-1+\frac{1}{3}\left(\frac{1}{3^{15}}\right)$$
$$=\ \frac{1}{3}\cdot\frac{1}{3^{15}}=\frac{1}{3^{15+1}}=\frac{1}{3^{16}}$$
Answer = D
Top
Post new topic Post Reply
• Page 1 of 1