Can You Find This Sum? | IMO Longlist

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  • Опубликовано: 12 дек 2024

Комментарии • 41

  • @vishalmittal934
    @vishalmittal934 2 года назад +45

    Another way: Multiply and divide by 4 - Write 4 as (4-0) for first term, (5-1) for second term and so on till (100-96) for last term. The series converts to telescoping sum and cancelling all terms leads to (1/4)*(97*98*99*100) for a really short answer.

    • @heliocentric1756
      @heliocentric1756 2 года назад +2

      Nice

    • @ramk4004
      @ramk4004 2 года назад +2

      Elegant

    • @satyapalsingh4429
      @satyapalsingh4429 2 года назад +2

      Elegant

    • @chikezienestor3394
      @chikezienestor3394 2 года назад

      Very smart strategy.

    • @leif1075
      @leif1075 2 года назад +1

      Why number 4 though? Seems random? What does that even mean 5-1?? There is no 5 in the second term or a 1..this doesn't make sense

  • @wesleysuen4140
    @wesleysuen4140 2 года назад +3

    The n-th term of such a sum is given by:
    n(n+1)(n+2)
    =[(n+1)(n+2)(n+3)(n+4)-n(n+1)(n+2)(n+3)]/4
    Then, as a telescopic sum, the sum of the first n terms is given by:
    =[(n+1)(n+2)(n+3)(n+4)-(1)(2)(3)(4)]/4

  • @willbishop1355
    @willbishop1355 2 года назад +4

    I used your first method, but slightly different. (Sum from n=2 to 98 of n^3) - (Sum from n=2 to 98 of n) = (Sum from n=1 to 98 of n^3) - 1 - (Sum from n=1 to 98 of n) + 1. So the 1s cancel out and it's just (Sum from n=1 to 98 of n^3) - (Sum from n=1 to 98 of n)
    = (98*99/2)^2 - (98*99/2)
    = (49*99)(49*99 - 1)
    = 4851 * 4850
    = 23527350

  • @two697
    @two697 2 года назад +15

    This was one of the easiest questions I've seen on this channel. Just use the formuals for the sum of n^3, n^2 and n, you'll get 97(98)(99)(100)/4

    • @fix5072
      @fix5072 2 года назад +2

      just imagine this being on the IMO, even I would've got points on that

    • @devroopsaha4020
      @devroopsaha4020 2 года назад +1

      That's why it's at the longlist and not the official question..

    • @two697
      @two697 2 года назад +1

      @@devroopsaha4020 the fact that this was even suggested is surprising

  • @dneary
    @dneary 2 года назад +1

    Since 1^3-1=0, you can extend both series to 1: \sum_{n=1}^{98}(n^3-n) works. There's also a calculus answer where you set f(x)=1+x+x^2+x^3+x^4+...+x^{99} = \frac{x^{100}-1}{x-1}, and calculate \lim_{x \to 1}f^(3)(x) using l'Hopital's rule.

  • @Stelios2711
    @Stelios2711 2 года назад +1

    Euler-Maclaurin summation formula is an overkill for that, but why not?

  • @suniltshegaonkar7809
    @suniltshegaonkar7809 2 года назад

    AT 3:02, you may write -6 in stead of -1, because you get 6 for the first term after simplification. any alternate method to verify?
    I cannot understand why -1 should be correct to apply the formula.

  • @ritam8767
    @ritam8767 2 года назад +2

    How on earth did this make the IMO longlist? It's just sum of n(n+1)(n+2) where n varies from 1 to 97. We already know the formulae for sum of n ^3, n^2 etc

  • @bosorot
    @bosorot 2 года назад

    5:00 . draw pascal triagle will visualize the proof very well.

  • @nirmankhan2134
    @nirmankhan2134 2 года назад

    Wow. What a coincidence. I thought of this problem and an way of solving it yesterday before going to sleep. 😆

  • @sarthakkumar8679
    @sarthakkumar8679 2 года назад +2

    JEE STUDENTS ARE U HERE

  • @skyhighbray313
    @skyhighbray313 2 года назад +1

    Am i the only one who did the summation of n=0 on bottom and 97 on top with the function being n^3+3n^2+2n?

  • @Aqua17292
    @Aqua17292 2 года назад +5

    How is this even in the IMO Shortlist 😂

  • @atpugnes
    @atpugnes 2 года назад

    How do you multiply so fast. Is it some mental math technique?

  • @jforjanexd4298
    @jforjanexd4298 Год назад

    Can you give me advice to a beginner like me? The team selection is in a month and I just started learning. And I just do easy Olympiad like AMO so I find it a bit difficult and I haven't learnt these

  • @ferramatis
    @ferramatis 2 года назад

    omg this is amazing

  • @mathematicalmonk1427
    @mathematicalmonk1427 2 года назад +2

    Interesting

  • @recepduzenli3263
    @recepduzenli3263 2 года назад +1

    Soru neyi soruyor hocam.sayilar yazılmış ama.neyi bulmamız isteniyor

  • @jordanmarx2584
    @jordanmarx2584 2 года назад

    I used a TI-84 Plus calculator to get the answer to this problem which is 23,527,350.

  • @admink8662
    @admink8662 2 года назад

    Falling power

  • @razer9728
    @razer9728 2 года назад

    Хорошее видно, самое странное что я все понял

  • @skyhighbray313
    @skyhighbray313 2 года назад +1

    Am i the only one who did the summation of n=0 on bottom and 97 on top with the function being n^3+3n^2+2n?