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Question regarding binomial theorem.

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data

    (√2 + 1)6 = I + f

    Where I is the sum of integer part of the expansion of (√2 + 1)6 and f is sum of the fraction part in (√2 + 1)6.

    2. Relevant equations

    (x+1)n = nC0 xn + nC1 xn-1 + nC2 xn-2 + ....... + nCn

    nCn = nC0 = 1

    3. The attempt at a solution

    I expanded (√2 + 1)6 , then simplified and then got the expression 44+99√2/2.
    Then I got I=44 which was not even the correct answer. The correct was 197. This question is a competitive level question.

    Thanks in advance... :smile:
     
  2. jcsd
  3. Jul 11, 2012 #2
    Hi sankalp :smile:

    The answer you get is incomplete. The second term also has an integral and fractional part itself. You need to add them to get your answer :wink:

    Another way to see it is that f is defined to be between 0 and 1, so is your fractional part between zero and one?


    PS : Your expansion itself seems incorrect to me. Recheck it.
     
  4. Jul 12, 2012 #3
    Hii Infinitum!! :smile:

    One way to do is to seriously find (√2 + 1)6. But that will be a noob way.

    Ok , so on expanding , I get :

    (√2 + 1)6 = 6C0 8 + 6C1 4√2 + 6C2 4 + 6C32√2 + 6C4 2 + 6C5 √2 + 6C6

    Now on solving , I get (√2 + 1)6 = 99+ 70√2
    I + f = 99+ 70√2

    Now what else can I do ? Any hint ?
     
  5. Jul 12, 2012 #4
    Yep. That sounds correct. Now you can use the approximate value of √2 to multiply, and hence get the integral part of the expansion.

    Hint : You only need to use 1.41 as your approximation, as any more digits will not change effect the integral part :wink:
     
  6. Jul 12, 2012 #5
    Did not recognize that this was so simple...

    99+ 70√2

    99+ 70(1.41)
    I = 99+98 = 197 !!

    Awesome!! :smile:

    Edit : All right , thanks for the efforts.....
     
    Last edited: Jul 12, 2012
  7. Jul 12, 2012 #6
    The method your professor uses can be applied in general to all such problems, so it is good in its own right. The one I suggest requires that you know the value of √2, which is frequently used and known. What about when it is √327? You would have to first find the square root, then approximate.
     
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