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

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New post Updated on: 11 Oct 2019, 09:01
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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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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.
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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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New post 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 ?

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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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New post 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
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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.
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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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New post 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 ?

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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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New post 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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Re: For a positive integer n, f(n) is defined as 1 + 1/2 + 1/3 + … + 1/n.  [#permalink]

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New post 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)
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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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