n = 2 ∏ ∞ n 3 + 1 n 3 − 1 If the value of above product is in the form b a , where a and b are coprime positive integers, find a + b .
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This is a misleading "solution", to put it politely. You make it look like ∏ n = 2 ∞ n + 1 n − 1 = 2 and ∏ n = 2 ∞ n 2 − n + 1 n 2 + n + 1 = 3 1 .
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Frankly speaking-Is that not so?? Can you please explain further why we cannot write it like that??
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No, it isn't so at all. Think about it: n + 1 n − 1 < 1 for all n , so, how could the product come out to be 2? You have to find the partial products ∏ n = 2 m n 3 + 1 n 3 − 1 , expressed in terms of m , and then take the limit as m goes to infinity.
The third line of your solution comes out to be 0 × ∞ , so, you cannot break the limit up like this.
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@Otto Bretscher – Can I edit my solution like this: First bracket is m ( m + 1 ) 2 while second one is 3 m 2 + m + 1 and when m → ∞ product is 3 2 .
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@Rishabh Jain – Right! As the end, write " P approaches 3 2 " to make a clean distinction between the partial product P and its limit.
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@Otto Bretscher – Done... Now its complete I think.... :)
You're posting fantastic solutions.Maybe you'll soon leave me behind in upvotes :P (+1)
Why can't we use compounendo and dividendo
Let's consider the general term a n ≡ n 3 + 1 n 3 − 1 = ( n + 1 ) ( n 2 − n + 1 ) ( n − 1 ) ( n 2 + n + 1 ) = ( n + 1 ) + n ( n − 1 ) ( n + 1 ) ( n − 1 ) + n ( n − 1 ) ( n + 1 ) = 1 + n ( n − 1 ) 1 1 + n ( n + 1 ) 1 . Let's call: x n ≡ 1 + n ( n − 1 ) 1 . We can write the original product as n = 2 ∏ ∞ a n = n = 2 ∏ ∞ x n x n + 1 . Numerators and denominators of consecutive terms therefore simplify. The partial product P N ≡ n = 2 ∏ N x n x n + 1 = x 2 x N + 1 = 1 + 2 ( 2 − 1 ) 1 1 + N ( N + 1 ) 1 = 3 2 [ 1 + N ( N + 1 ) 1 ] . Thus N → ∞ lim P N = 3 2 .
Consider:
n = 2 ∏ ∞ n 3 + 1 n 3 − 1 = n = 2 ∏ m ( n + 1 ) ( n 2 − n + 1 ) ( n − 1 ) ( n 2 + n + 1 ) = 3 ˙ 3 1 ˙ 7 ˙ 4 ˙ 7 2 ˙ 1 3 ˙ 5 ˙ 1 3 3 ˙ 2 1 ˙ 6 ˙ 2 1 4 ˙ 3 1 ˙ 7 ˙ 3 1 5 ˙ 4 3 ˙ . . . = 3 2
⇒ a + b = 2 + 3 = 5
An old classic Putnam problem......Question A1 from the '77 Exam!
Very popular problem indeed. I have seen this a couple of time on Brilliant only.
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Using a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) P = n = 2 ∏ m ( n + 1 ) ( n 2 − n + 1 ) ( n − 1 ) ( n 2 + n + 1 ) = ( n = 2 ∏ m n + 1 n − 1 ) × ( n = 2 ∏ m n 2 − n + 1 n 2 + n + 1 ) = ( 3 1 × 4 2 × 5 3 × 6 4 × . . . . m + 1 m − 1 ) × ( 3 7 × 7 1 3 × 1 3 2 1 × 2 1 3 1 × . . . m 2 − m + 1 m 2 + m + 1 ) = 3 ( m ) ( m + 1 ) 2 ( m 2 + m + 1 ) When m → ∞ , P approaches 3 2
Hence, 2 + 3 = 5