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

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Math Revolution GMAT Instructor
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For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

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Updated on: 11 Oct 2019, 09:01
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65% (hard)

Question Stats:

57% (02:34) correct 43% (02:29) wrong based on 46 sessions

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[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 $$10+f(1)+f(2)+…+f(9)$$?

$$A. f(11)$$

$$B. 9f(9)$$

$$C. 10f(10)$$

$$D. f(10)$$

$$E. 22$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" 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 Joined: 20 Jul 2017 Posts: 1076 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 09 Oct 2019, 10:09 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 Joined: 16 Aug 2015 Posts: 8141 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 Updated on: 11 Oct 2019, 09:00 1 => $$10 + f(1) + f(2) + … + f(9)$$ $$= 10 + ( 1 ) + ( 1 + \frac{1}{2} ) + ( 1 + \frac{1}{2} + \frac{1}{3} ) + … + ( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$ $$= 10 + 1*9 + (\frac{1}{2})*8 + (\frac{1}{3})*7 + … + (\frac{1}{8})*2 + (\frac{1}{9})$$ $$= 10 + 1*(10 – 1) + (\frac{1}{2})*(10-2) + (\frac{1}{3})*(10-3) + … + (\frac{1}{8})*(10-8) + (\frac{1}{9})(10-9)$$ $$= 10 + (-1) + (-1) + … + (-1) + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} )$$ $$= 1 + 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9})$$ $$= 10( 1 + \frac{1}{2} + \frac{1}{3} + … + \frac{1}{9} + \frac{1}{10})$$ $$= 10f(10)$$ Therefore, C is the answer. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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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
Joined: 20 Jul 2017
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Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

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11 Oct 2019, 01:34
MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

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

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

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

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

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

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

$$= f(10)$$

Therefore, D is the answer.
Answer: D

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
Joined: 16 Aug 2015
Posts: 8141
GMAT 1: 760 Q51 V42
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For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

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11 Oct 2019, 09:01
Dillesh4096 wrote:
MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

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

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

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

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

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

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

$$= f(10)$$

Therefore, D is the answer.
Answer: D

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.
Sorry about the mistake.
The answer is C.
The official answer is corrected.
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Joined: 25 Jan 2013
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Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

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20 Oct 2019, 10:15
MathRevolution wrote:
=>

$$10 + f(1) + f(2) + … + f(9)$$

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

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

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

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

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

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

$$= 10f(10)$$

Therefore, C is the answer.

Answer: C

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

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