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

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

gmatophobia · Answer

10+f(1)+f(2)+…+f(9)

10+(\frac{1}{1}) + (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3}) + .... + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} ... \frac{1}{8})+ (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} ... \frac{1}{8} + \frac{1}{9})

10+(\frac{1}{1} * 9 ) + (\frac{1}{2} * 8) + (\frac{1}{3} * 7) + .... + (\frac{1}{8}*2) + (\frac{1}{9}*1)

\approx 10 + 9 + 4 + 2 + .... > 25.XXX

So, we can conclude that the sum of the series is a value greater than 25.

At this stage, we can eliminate A, D, and E as the values represented by those options are less than 25.

Let's take a look at option B

9*f(9) = 9 * (1 + \frac{1}{2} + \frac{1}{3} + ... \frac{1}{9})

=  9 * (1 + 0.5 + 0.33 + 0.25 + 0.5 + 0.16 + 0.14 + 0.12 + 0.11) \approx 2.8

=  9 * 2.8 \approx 25

However, we know that the answer should exceed 25. Hence we can eliminate B.

Option C

We can also prove Option C, by finding the approximate value of 10f(10)

=  9 * (1 + 0.5 + 0.33 + 0.25 + 0.5 + 0.16 + 0.14 + 0.12 + 0.11 + 0.10) \approx (2.8 + 0.10) \approx 2.9

=  10 * 2.9 \approx 29

This sum seems more realistic than the sum in Option B

Option C

Theprince · Answer

We can rule out f(10) and f(11) because we are adding 10 to all the terms so definetly it will be more than these ans choices.
we can rule out 22 as rarely these fractions will give us integer.

Now for last 2 options we have 9*f(9) and 10*f(10) which means we have to convert all the terms to only one function f(9) or f(10).

by looking at the terms we know for f(9) to covert into f(10) we need  \frac{1}{10} . similary for f(8) to convert it into f(10) we need  \frac{1}{9} + \frac{1}{10}

if we do this for all numbers then we will need  \frac{1}{10}  -->10 times,  \frac{1}{9}  -->9 times,  \frac{1}{8} --> 8 times ..and finally  \frac{1}{1} --> 1 time..addtion of this can give us 10 ...

final ans  C

jm2k5 · Answer

f(1) = 1 
 f(2) = 1+\frac{1}{2} 
 f(3) = 1+\frac{1}{2}+\frac{1}{3} 
.
.
.
.
 f(9) = 1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{8}+\frac{1}{9} 
_______________________________________________________________________________________________________

f(1)+f(2)+f(3)+......+f(9) = 9*1+8*\frac{1}{2}+7*\frac{1}{3}+6*\frac{1}{4}+.........+2*\frac{1}{8}+\frac{1}{9} 
=  9*1+(9-1)*\frac{1}{2}+(9-2)*\frac{1}{3}+(9-3)*\frac{1}{4}+.........+(9-7)*\frac{1}{8}+(9-8)*\frac{1}{9} 
=  (9*1+9*\frac{1}{2}+9*\frac{1}{3}+9*\frac{1}{4}+.........+9*\frac{1}{8}+9*\frac{1}{9})-(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+....+\frac{8}{9})

=  9*f(9) -

=  10*f(9) - 9

Adding 10 on both sides,

f(1)+f(2)+f(3)+......+f(9) + 10 = 10*f(9) - 9+10

or,  f(1)+f(2)+f(3)+......+f(9) + 10 = 10*f(9) +1

or,  f(1)+f(2)+f(3)+......+f(9) + 10 = 10*

Gemmie · Answer

f(1) = 1

f(2) =  1 + \frac{1}{2}

f(3) =  1 + \frac{1}{2} + \frac{1}{3} 
...
f(9) =  1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{9}

f(1) + f(2) + ... + f(9) =  9 + (8 * \frac{1}{2}) + (7 * \frac{1}{3}) + ... + (2 * \frac{1}{8}) + \frac{1}{9}

Check f(11) =   1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{11} 
==> too small

Check f(10)  => similar to f(11)

Check 9 * f(9)  =  9 + \frac{9}{2} + \frac{9}{3} + ... + \frac{9}{9}

9 * f(9) - (f(1) + ... + f(9))

= (\frac{9}{2} - \frac{8}{2}) + (\frac{9}{3} - \frac{7}{3}) + ... + (\frac{9}{9} - \frac{8}{9})

=  \frac{1}{2} + \frac{2}{3} + ... + \frac{8}{9}  
=> not 10

Check 10 * f(10) 
 10 * f(10) = 10 + 10 * \frac{1}{2} + 10 * \frac{1}{3} + ... + 10 * \frac{1}{9} + 10 * \frac{1}{10}

10 * f(10) - (f(1) + ... + f(9))

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

= 1+ \frac{2}{2} + \frac{3}{3} + ... + \frac{10}{10}

= 1 + 9
= 10

=> 10 * f(10) = 10 + f(1) + f(2) + ... + f(9)

BrushMyQuant · Answer

f(10)  =  1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{9} + \frac{1}{10}   = f(9) +  \frac{1}{10} 
=> f(9) = f(10) -  \frac{1}{10}

Similarly
 f(8)  = f(10) -  \frac{1}{10}  -  \frac{1}{9} 
 f(7)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  
  f(6)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  
  f(5)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  -  \frac{1}{6}  
  f(4)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  -  \frac{1}{6}  -  \frac{1}{5} 
  f(3)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  -  \frac{1}{6}  -  \frac{1}{5}  -  \frac{1}{4} 
  f(2)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  -  \frac{1}{6}  -  \frac{1}{5}  -  \frac{1}{4}  -  \frac{1}{3} 
  f(1)  = f(10) -  \frac{1}{10}  -  \frac{1}{9}  -  \frac{1}{8}  -  \frac{1}{7}  -  \frac{1}{6}  -  \frac{1}{5}  -  \frac{1}{4}  -  \frac{1}{3}  -  \frac{1}{2}

=>  f(1) + f(2) + f(3)... f(9)  = 9 * f(10) - 9 * \frac{1}{10}  - 8 *  \frac{1}{9}  - 7 *  \frac{1}{8}  - 6 *  \frac{1}{7}  - 5 *  \frac{1}{6}  - 4 *  \frac{1}{5}  - 3 *  \frac{1}{4}  - 2 *  \frac{1}{3}  - 1 *  \frac{1}{2}

Adding and subtracting f(10) on the right hand side 
 => f(1) + f(2) + f(3)... f(9) = 10 * f(10) - f(10) -  \frac{9}{10}  -  \frac{8}{9}  -  \frac{7}{8}  -  \frac{6}{7}  -  \frac{5}{6}  -  \frac{4}{5}  -  \frac{3}{4}  -  \frac{2}{3}  -  \frac{1}{2}

Substituting value of -f(10)  = -( 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{9} + \frac{1}{10}  ) =  - 1 - \frac{1}{2} - \frac{1}{3} - ... - \frac{1}{9} - \frac{1}{10}

=>  f(1) + f(2) + f(3)... f(9)  = 10 * f(10)  - 1 - \frac{1}{2} - \frac{1}{3} - ... - \frac{1}{9} - \frac{1}{10}   -  \frac{9}{10}  -  \frac{8}{9}  -  \frac{7}{8}  -  \frac{6}{7}  -  \frac{5}{6}  -  \frac{4}{5}  -  \frac{3}{4}  -  \frac{2}{3}  -  \frac{1}{2}  
 =>  f(1) + f(2) + f(3)... f(9)  = 10*f(10) - 1 -  \frac{10}{10}  -  \frac{9}{9}  -  \frac{8}{8}  -  \frac{7}{7}  -  \frac{6}{6}  -  \frac{5}{5}  -  \frac{4}{4}  -  \frac{3}{3}  -  \frac{2}{2} 
 =>  f(1) + f(2) + f(3)... f(9)  = 10*f(10) - 10
=> 10 + f(1) + f(2) + f(3)... f(9) = 10*f(10) 
 
So,  Answer will be C 
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters

https://www.youtube.com/watch?v=b7JoMX0gk8M

Kinshook · Answer

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) ?

f(1) = 1 
f(2) = 1 + 1/2
f(3) = 1 + 1/2 + 1/3

...
f(9) = 1 + 1/2 + 1/3 + .... + 1/9

10+f(1)+f(2)+...+f(9) = 10 + 9*1 + 8*\frac{1}{2} + 7*\frac{1}{3} + 6*\frac{1}{4} + 5*\frac{1}{5} + 4*\frac{1}{6} + 3*\frac{1}{7} + 2*\frac{1}{8} + 1*\frac{1}{9 }
= 10 + \frac{10}{1} - 1 + \frac{10}{2} - 1 + \frac{10}{3} - 1 + \frac{10}{4} - 1 + \frac{10}{5} - 1 + \frac{10}{6} - 1 + \frac{10}{7} - 1 + \frac{10}{8} - 1 + \frac{10}{9} - 1
= 1 + 10(1 + \frac{1}{2} + .... + \frac{1}{9} ) = 10(1 + \frac{1}{2} + .... + \frac{1}{10}) = 10f(10)

Since  \frac{(10-n)}{n} = \frac{10}{n} - 1

IMO C

abhikc3004 · Answer

Crazy question this was, how do you solve this in 2min?

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

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