GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 13 Nov 2019, 21:45 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.

Author Message
TAGS:

### Hide Tags

Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8131
GMAT 1: 760 Q51 V42 GPA: 3.82
For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

1
6 00:00

Difficulty:   65% (hard)

Question Stats: 58% (02:34) correct 42% (02:29) wrong based on 45 sessions

### HideShow timer Statistics

[GMAT math practice question]

For a positive integer $$n, f(n)$$ is defined as $$1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{n}.$$ What is the value of $$10+f(1)+f(2)+…+f(9)$$?

$$A. f(11)$$

$$B. 9f(9)$$

$$C. 10f(10)$$

$$D. f(10)$$

$$E. 22$$

_________________

Originally posted by MathRevolution on 09 Oct 2019, 07:30.
Last edited by MathRevolution on 11 Oct 2019, 09:01, edited 1 time in total.
VP  D
Joined: 20 Jul 2017
Posts: 1055
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

MathRevolution wrote:
[GMAT math practice question]

For a positive integer $$n, f(n)$$ is defined as $$1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{n}.$$ What is the value of $$10+f(1)+f(2)+…+f(9)$$?

$$A. f(11)$$

$$B. 9f(9)$$

$$C. 10f(10)$$

$$D. f(10)$$

$$E. 22$$

$$f(n)$$ = $$1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{n}.$$

So, $$f(n+1)$$ = $$1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{(n+1)}.$$

—> $$f(n+1)$$ = $$f(n) +\frac{1}{(n+1)}.$$

—> $$(n+1)f(n+1) = (n+1)f(n) + 1$$

10*f(10) = 10*f(9) + 1
9*f(9) = 9*f(8) + 1
8*f(8) = 8*f(7) + 1
.
.
3*f(3) = 3*f(2) + 1
2*f(2) = 2*f(1) + 1 = f(1) + 2 [Since f(1) = 1]

Adding all, 10*f(10) = f(9) + f(8) + f(7) + . . . . + f(1) + 10

IMO Option C

Hi MathRevolution,

Can you let me know where I’m going wrong ?

Posted from my mobile device
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8131
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

1
=>

$$10 + f(1) + f(2) + … + f(9)$$

$$= 10 + ( 1 ) + ( 1 + \frac{1}{2} ) + ( 1 + \frac{1}{2} + \frac{1}{3} ) + … + ( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 10 + 1*9 + (\frac{1}{2})*8 + (\frac{1}{3})*7 + … + (\frac{1}{8})*2 + (\frac{1}{9})$$

$$= 10 + 1*(10 – 1) + (\frac{1}{2})*(10-2) + (\frac{1}{3})*(10-3) + … + (\frac{1}{8})*(10-8) + (\frac{1}{9})(10-9)$$

$$= 10 + (-1) + (-1) + … + (-1) + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 1 + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9})$$

$$= 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} + \frac{1}{10})$$

$$= 10f(10)$$

_________________

Originally posted by MathRevolution on 11 Oct 2019, 01:07.
Last edited by MathRevolution on 11 Oct 2019, 09:00, edited 1 time in total.
VP  D
Joined: 20 Jul 2017
Posts: 1055
Location: India
Concentration: Entrepreneurship, Marketing
WE: Education (Education)
Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

$$= 10 + ( 1 ) + ( 1 + \frac{1}{2} ) + ( 1 + \frac{1}{2} + \frac{1}{3} ) + … + ( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 10 + 1*9 + (\frac{1}{2})*8 + (\frac{1}{3})*7 + … + (\frac{1}{8})*2 + (\frac{1}{9})$$

$$= 10 + 1*(10 – 1) + (\frac{1}{2})*(10-2) + (\frac{1}{3})*(10-3) + … + (\frac{1}{8})*(10-8) + (\frac{1}{9})(10-9)$$

$$= 10 + (-1) + (-1) + … + (-1) + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 1 + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9})$$

$$= 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} + \frac{1}{10})$$

$$= f(10)$$

The highlighted part is 10*f(10)
So, answer has to be C
Bunuel Can you pls edit the OA ?

Posted from my mobile device
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8131
GMAT 1: 760 Q51 V42 GPA: 3.82
For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

Dillesh4096 wrote:
MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

$$= 10 + ( 1 ) + ( 1 + \frac{1}{2} ) + ( 1 + \frac{1}{2} + \frac{1}{3} ) + … + ( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 10 + 1*9 + (\frac{1}{2})*8 + (\frac{1}{3})*7 + … + (\frac{1}{8})*2 + (\frac{1}{9})$$

$$= 10 + 1*(10 – 1) + (\frac{1}{2})*(10-2) + (\frac{1}{3})*(10-3) + … + (\frac{1}{8})*(10-8) + (\frac{1}{9})(10-9)$$

$$= 10 + (-1) + (-1) + … + (-1) + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 1 + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9})$$

$$= 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} + \frac{1}{10})$$

$$= f(10)$$

The highlighted part is 10*f(10)
So, answer has to be C
Bunuel Can you pls edit the OA ?

Posted from my mobile device

You are right.
_________________
Intern  B
Joined: 25 Jan 2013
Posts: 29
Location: United States
Concentration: General Management, Entrepreneurship
Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

### Show Tags

MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

$$= 10 + ( 1 ) + ( 1 + \frac{1}{2} ) + ( 1 + \frac{1}{2} + \frac{1}{3} ) + … + ( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 10 + 1*9 + (\frac{1}{2})*8 + (\frac{1}{3})*7 + … + (\frac{1}{8})*2 + (\frac{1}{9})$$

$$= 10 + 1*(10 – 1) + (\frac{1}{2})*(10-2) + (\frac{1}{3})*(10-3) + … + (\frac{1}{8})*(10-8) + (\frac{1}{9})(10-9)$$

$$= 10 + (-1) + (-1) + … + (-1) + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$

$$= 1 + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9})$$

$$= 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} + \frac{1}{10})$$

$$= 10f(10)$$

Can anyone help me understand below lines. How it was simplified.

=10+1∗(10–1)+(12)∗(10−2)+(13)∗(10−3)+…+(18)∗(10−8)+(19)(10−9)=10+1∗(10–1)+(12)∗(10−2)+(13)∗(10−3)+…+(18)∗(10−8)+(19)(10−9)

=10+(−1)+(−1)+…+(−1)+10(1+12+13+…+19) Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.   [#permalink] 20 Oct 2019, 10:15
Display posts from previous: Sort by

# For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  