Jump To Content

LearnHub

mahesh25

Person

mahesh25

Primary Navigation

14 Comments

  1. From the lesson Permutations and Combinations

    Sat, 11 Jul 2009 14:07:07 -0000

    IN THE FOLLOWING LESSON IAM GIVING THE COMPLETE LESSON

    http://maths.learnhub.com/lesson/13338-permutations-and-combinations

    Rating
    1
    Rate Up

  2. From the question Maths JEE

    Tue, 07 Jul 2009 10:24:44 -0000

    the answer is 2 first solve the given equation for x Then after finding the value of x u substitute that value in the expression which we want find

    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

  3. From the question Maths JEE

    Tue, 07 Jul 2009 10:22:39 -0000

    the answer is 2

    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

  4. From the question Maths JEE

    Tue, 07 Jul 2009 10:00:31 -0000

    please mention ur problem correctly i can not understand ur problem ok

    Vote
    Current Rating
    1
    Rate Up
    Rate Down
    1 Total Vote

  5. From the question FUNCTION -JEE

    Tue, 07 Jul 2009 09:39:51 -0000

    f(n) = \frac{2.2005}{n.(n+1)}

    This formulae is useful to solve the problem

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    2 Total Votes

  6. From the question FUNCTION -JEE

    Sat, 04 Jul 2009 11:46:54 -0000

    Sorry friend the above solution is wrong

    The correct solution is here

    given that f(1)=2005

    and for n > 1 f(1) + f(2) + f(3) + ....... +f(n) + = n^2 f(n) .......... (1)

    Equation 1 can be written as

    \frac{ f(1) + f(2) + f(3) + .......+f(n-1) }{} +f(n) + = n^2 ... (2)

    Now \frac{f(1) +f(2) +f(3) + .......+f(n-1) }{} = (n-1)^2 f(n-1).... (3)

    ( Since from equation 1)

    Now substitute the vale of equation 3 in equation 2

    We get (n-1)^2 f(n-1) +f(n)= n^2 f(n)

    This can be written as (n-1)(n-1) .f(n-1) = n^2 .f(n) – f(n)

    (n-1)(n-1) .f(n-1) = (n^2 -1 ) .f(n)

    (n-1)(n-1) .f(n-1) = (n -1)(n+1) .f(n)

    (n-1).f(n-1) = (n+1).f(n)

    \frac{f(n)}{f(n-1)} = \frac{(n-1)}{(n+1)} ......... (4)

    now given that f(1)=2005

    Now to find f(2) put n =2 in equation (4)

    We get \frac{f(2)}{f(1)} = \frac{1}{3}

    f(2)=\frac{1.f(1)}{3} =\frac{1.2005}{3} ( since f(1) =2005 )

    Now to find f(3) put n =3 in equation (4)

    We get \frac{f(3)}{f(2)} = \frac{2}{4}

    f(3)=\frac{2.f(2)}{4} =\frac{2}{4} .\frac{1.2005}{3} ( since f(2) =\frac{1.2005}{3} )

    Now to find f(4) put n =4 in equation (4)

    We get \frac{f(4)}{f(3)} = \frac{3}{5}

    f(4)=\frac{3.f(3)}{5} =\frac{3}{5} .\frac{2}{4} .\frac{1.2005}{3} ( since f(3)==\frac{2}{4} .\frac{1.2005}{3}

    Now to find f(5) put n =5 in equation (4)

    We get \frac{f(5)}{f(4)} = \frac{4}{6}

    f(5)=\frac{4.f(4)}{6} =\frac{4}{6} . \frac{3}{5} .\frac{2}{4} .\frac{1.2005}{3} ( since f(4)= \frac{3}{5} .\frac{2}{4} .\frac{1.2005}{3}

    f(5) = \frac{4}{6} . \frac{3}{5} .\frac{2}{4} .\frac{1.2005}{3}

    This is is true for first 5 terms

    The above equation can be written as

    f(5) = \frac{4}{6} . \frac{3}{5} .\frac{2}{4} .\frac{1}{3}.\frac{1.2005.2}{2}

    If we generalise for n terms

    The numarator is 1.2.3.4…....(n-1) = (n-1)!

    And the denominator is 1.2.3.4.5.6…..( n+1) = (n+1)! ( by 0bserving the above statement )

    so f(n) = \frac{(n-1)....4.3.2.1.2005.2}{(n+1).n........6.5.4.3.2.1} = \frac{2.(n-1)!.2005}{(n+1)!}

    f(n) = \frac{2.(n-1)!.2005}{(n+1).n.(n-1)!}

    f(n) = \frac{2.2005}{n.(n+1)}

    Now substitute n = 2004 in the above equation

    We have f(2004) = \frac{2.2005}{2004.2005}

    f(2004) = \frac{2}{2004}
    f(2004) = \frac{1}{1002}
    Actions
    • 1 Reply
    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

    1. ANKIT PANDE said – Tue, 07 Jul 2009 09:40:14 -0000

      YOUR ASWER IS CORRECT, YOUR METHOD IS CORRECT BUT THIS SOLUTION WOULD TAKE MORE THEN 2 MINUTES TO BE WORKED OUT ,THIS MEANS IF THIS QUESTION COMES ON JEE YOU WILL LOOSE TWO TO THREE QUESTION. MOREOVER WE ARE LOOKING FOR SMALLEST EASIEST SOLUTION .THEREFORE THIS SOLUTION CAN’T BE TAKEN NOR CAN BE THOUGHT OF POLLING AS THE BEST SOLUTION.

      Rating
      2
      Rate Up
      Author
      ANKIT PANDE
      ANKIT PANDE
      Authority 80

  7. From the question FUNCTION-JEE

    Fri, 03 Jul 2009 10:13:54 -0000

    given that f(1)=2005

    f(1) + f(2) + f(3) +...............+f(n)=n^2 .f(n) for all n>1 .................(1)

    This means for n=1 f(1)=2005 and for the above values > 1 i.e n=2,3,4….... the values can be found through the equation 1

    to find f(2) put n= 2 in equation 1 we get

    Actions
    • 1 Reply
    Vote
    Current Rating
    -2
    Rate Up
    Rate Down
    2 Total Votes

    Post Comments

    1. Sureshbala said – Mon, 06 Jul 2009 08:02:55 -0000

      Hi,

      This has been answered…Check my reply

      Rating
      1
      Rate Up
      Author
      Sureshbala
      Sureshbala
      Authority 787

  8. From the lesson SHORTCUT METHOD TO FIND RANK OF A GIVEN WORD

    Tue, 02 Dec 2008 08:08:06 -0000

    ya it is applicable when all the given things are not repeated

    Rating
    0
    Rate Up

  9. From the presentation Uses of Mathematics in daily Life

    Wed, 12 Nov 2008 12:40:25 -0000

    good lesson

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

  10. From the lesson SHORTCUT METHOD TO FIND RANK OF A GIVEN WORD

    Wed, 29 Oct 2008 13:42:51 -0000

    Sorry I donot have any Idea abt ur question. I think of it.But I donot get any idea. If U have any Shortcut method please publish in my community.

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

  11. From the lesson Permutations and Combinations

    Wed, 29 Oct 2008 13:29:50 -0000

    Thank you Mr.sharma I correct that mistake.

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

  12. From the presentation LINEAR EQUATIONS IN TWO VARIABLES

    Mon, 27 Oct 2008 12:03:39 -0000

    Nice lesson.Please explain above concepts with an example.Then it is easily to understand.

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

  13. From the lesson Math Lecture Videos & Lessons

    Sun, 26 Oct 2008 12:55:53 -0000

    nice

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes

  14. From the lesson Equation Editor

    Mon, 06 Oct 2008 10:11:15 -0000

    how can i strike a letter

    Vote
    Current Rating
    0
    Rate Up
    Rate Down
    No Votes