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.
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
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
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.
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.
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
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
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.
Nice
Elegant
Elegant
Very smart strategy.
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
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
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
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
just imagine this being on the IMO, even I would've got points on that
That's why it's at the longlist and not the official question..
@@devroopsaha4020 the fact that this was even suggested is surprising
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.
Euler-Maclaurin summation formula is an overkill for that, but why not?
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.
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
yes i was too shocked
5:00 . draw pascal triagle will visualize the proof very well.
Wow. What a coincidence. I thought of this problem and an way of solving it yesterday before going to sleep. 😆
JEE STUDENTS ARE U HERE
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?
How is this even in the IMO Shortlist 😂
Ikr
Longlist
How do you multiply so fast. Is it some mental math technique?
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
omg this is amazing
Interesting
Soru neyi soruyor hocam.sayilar yazılmış ama.neyi bulmamız isteniyor
I used a TI-84 Plus calculator to get the answer to this problem which is 23,527,350.
Falling power
Хорошее видно, самое странное что я все понял
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?