Can you sum it?
r = 3 ∏ ∞ ( r 3 + 8 r 3 − 8 ) = ? Give your answer up to 3 decimal places.
The answer is 0.285.
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1 solution
Moderator note:
Can you explain how "after cancelling ..."?
Can you explain how "after cancelling ..."?
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5 . 6 . 7 . . . . . . . . ∞ 1 . 2 . 3 . 4 . . . . . . . ∞ - In the sequence the next terms in the numerator and denominator both cancel off. in the next sequence , 7 . 1 2 . 1 9 . 2 8 . 3 9 . 5 2 . . . . . . ∞ 1 9 . 2 8 . 3 9 . 5 2 . . . . . . ∞ = 7 . 1 2 1 so everything gets cut off after that. Thus the answer = 7 . 1 2 1 . 2 . 3 . 4 = 7 2
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How do we know that "everything gets cut off after that"? What is the explanation for that happening, other than observation?
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@Calvin Lin – As far as I conclude, In case of the first fraction, For the fact that (r-2) & (r+2) for any integer r is that (r+2)-(r-2) = 4 So (r+2) comes as the 4th number from (r-2) , Naturally for any r , (r-2) occupies the value of (r+2) if r is increased by 4. So after 4 terms the numerator gets equal to the denominator. For the Next case,
(
r
−
1
)
2
+ 3 &
(
r
+
1
)
2
+ 3 are the numerator and denominator respectively.
(r+1)-(r-1) = 2 , so after 2 terms (r-1) obtains the value of (r+1).
So after 2 terms (r-1) will take the values of (r+1) and the constant term is same '3'.
So the denominator gets the values same as the numerator.
So they cancel of after 2 terms.
This is what I understood from the fact.
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@Aditya Narayan Sharma – Great! Please add that explanation in. You can refer to the telescoping series - product for ideas on how to clearly demonstrate this.
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@Calvin Lin – Definitely Sir . i hope i was able to explain.
r = 3 ∏ ∞ ( r 3 + 8 r 3 − 8 ) Here We should factorise them and write seperately, r = 3 ∏ ∞ ( r + 2 r − 2 r 2 − 2 r + 4 r 2 + 2 r + 4 )
= 5 . 6 . 7 . . . . . . . . ∞ 1 . 2 . 3 . 4 . . . . . . . ∞ 7 . 1 2 . 1 9 . 2 8 . . . . . . . . ∞ 1 9 . 2 8 . . . . . . . ∞ [Putting the values of r for first few numbers]
For the fact that (r-2) & (r+2) for any integer r is that (r+2)-(r-2) = 4 So (r+2) comes as the 4th number from (r-2) , Naturally for any r , (r-2) occupies the value of (r+2) if r is increased by 4. So after 4 terms the numerator gets equal to the denominator. For the Next case,
( r − 1 ) 2 + 3 & ( r + 1 ) 2 are the numerator and denominator respectively. (r+1)-(r-1) = 2 , so after 2 terms (r-1) obtains the value of (r+1). So after 2 terms (r-1) will take the values of (r+1) and the constant term is same '3'. So the denominator gets the values same as the numerator. So they cancel of after 2 terms. After Cancellling We are left with 7 2
So the Answer is : 0.285 (approx)