Monday, August 15, 2016

கண்டுபிடி!

$$\frac{200^2+199^2+198^2+197^2+\dots+2^2+1^2}{200^2-199^2+198^2-197^2+\dots+2^2-1^2}$$

13 comments:

  1. தொகுதி -பின்னத்தில் மேலே இருப்பது
    1^2 + 2^2 + ... + 200^ = n (n+1)(2n+1)/6 = (200 x 201 x 401)/6 =2686700 ...... (1)
    இப்பொழுது,
    1^2 + 3 ^2 + ... 199^2 = n (2n+1) (2n-1)/3 = (100x201x199)/6 = 1333300 .... (2)

    ஆக, பகுதி -பின்னத்தில் கீழே இருப்பது (1) - (2)= 1333400

    ஆக, பின்னத்தின் மதிப்பு 2686700/1353400
    =1.985 1485 1485 ...
    சரியான விடை மாதிரித் தெரியவில்லை :)

    சரவணன்

    Saravanan

    =

    ReplyDelete
    Replies
    1. In the denominator, alternate signs are negative.

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    2. Yes, that's why I subtracted the sum of all odd terms [which is 1333300] from the sum of the whole series and got 1353400 for the denominator. For the numerator I have 2686700, hence got the fraction as 2686700/1353400 = 1.985 1485 1485...
      Saravanan

      Delete
  2. Kouchik is right. Sudhakar, you are wrong with your answer. The formula you have posted which I have not approved is slightly faulty. I can see how you have worked out, but you need to check your n & 2n.

    ReplyDelete
    Replies
    1. (2n+1)/3 if n is odd then -ve, is not it correct. n=200, (2*200+1) = 401/3=133.6667

      Delete
    2. (2n+1)/3 is correct. But you need to define "n" correctly. You wouldn't have gotten 200/3 as answer then.

      Delete
  3. I think the answer is 1.

    ReplyDelete
  4. N = 200*201*401/6
    x2 - (x-1)2 = 2x-1
    D = 2(200+198+..+2) - 100

    ReplyDelete
  5. (200*201*401/6)/((200*201*101/3)-(100*201*199/3))
    =2686700/(1353400-1333300)
    =2686700/20100
    =133.666667

    ReplyDelete