Can you spot the trick?
R = x = 1 ∑ ∞ x 8 − 3 x 6 + 2 7 x 2 − 8 1 x 5 − 9 x
If R can be written in the form q p , where p and q are coprime positive integers , find ∣ p − q ∣ .
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3 solutions
Yep, cool question(reshared and liked), cooolest solution!!(upvoted)
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Oh.. Thank you very much... Keep solving and posting... :-)
Nice! problem and solution! :)
Could you explain how you calculated that the sum of the telescopic series u got at the very end is 1+1+1/3?
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Write some of the starting terms and you will see how the series telescopes... ( 1 − 7 1 ) + ( 1 − 1 3 1 ) + ( 3 1 − 2 1 1 ) + ( 7 1 + 3 1 1 ) + ( 1 3 1 − 4 3 1 ) ⋯
See above the pattern how the terms cancel.All except first and last three terms cancel. Last three terms approach zero as x → ∞ so only first three terms remain and that's what I've written.. :-)
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@Rishabh Cool
There is no need to calculate the terms, you can show telescopy without calculating the terms.
Notice that, x 2 + 3 x + 3 = ( x + 3 ) 2 − 3 ( x + 3 ) + 3 .
Now, 6 1 [ n = 1 ∑ ∞ ( x 2 − 3 x + 3 1 − x 2 + 3 x + 3 1 ) ] = 6 1 [ n = 1 ∑ ∞ x 2 − 3 x + 3 1 − n = 1 ∑ ∞ ( x + 3 ) 2 − 3 ( x + 3 ) + 3 1 ] = 6 1 [ n = 1 ∑ ∞ x 2 − 3 x + 3 1 − n = 4 ∑ ∞ x 2 − 3 x + 3 1 ] = 6 1 [ 1 + 1 + 3 1 ] = 1 8 7 .
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@Aditya Sky – Nice I tried it this way but could not get it. Thanks for teaching me. up voted.
Thanks a lot :)
Nice, well done.
Plzz tell how we came to the second step of factorising the denominator
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(Basically its the Sophie Germain identity)
x 4 − 3 x 2 + 9 = ( x 4 + 6 x 2 + 9 ) − 9 x 2 = ( x 2 + 3 ) 2 − ( 3 x ) 2 = ( x 2 + 3 − 3 x ) ( x 2 + 3 + 3 x )
I also got the solution as 1/6 (2+(1/3)).But then i simplify it as 1/6 8/3=4/9=p/q. so absolute value of p-q =5.A silly mistake!...
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No problem.... You can always learn from them... :-)
You are kidding with me, the sum is 0.4924111
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No.... The answer is indeed correct
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sorry, there was a mistake in my Pascal sequence
sorry, it was a mistake in my Pascal sequence
For those yet unfamiliar with telescopic series, here is a more laborious solution:
Simplify the fraction to get: x 2 ( x 2 − 3 ) + 9 x .
Then do the summation manually until x = 2 0 , or using your calculator until x = 5 0 0 or so.
It will be clear that the series converges to 0 . 3 8 8 8 8 8 8 . . . .
Now all we need to do is to convert that into a rational fraction.
Let x = 0 . 3 8 8 8 8 8 8 . . . .
Then 1 0 x = 3 . 8 8 8 8 8 8 . . . and 1 0 0 x = 3 8 . 8 8 8 8 8 . . . .
Subtract the last two to get 9 0 x = 3 5 , or x = 1 8 7
So, the answer is the absolute value of 7 − 1 8 , i.e. 1 1 .
This is not a good solution. No finite number of terms will allow you to know with certainty that it converges to 7/18. You can, with only minor modifications, come up with series which will sum to 0.388888888889.
I know. I never said it was a good solution. But it sure helps in making a guess!
I solved it the Rishabh Cool way with long way for telescopic series.
Numerator : x ( x 4 − 9 )
Also, x 6 + 2 7 = ( x 2 + 3 ) ( x 4 − 3 x 2 + 9 ) = ( x 2 + 3 ) ( x 2 + 3 x + 3 ) ( x 2 − 3 x + 3 )
Hence Denominator : ( x 4 − 9 ) ( x 2 − 3 x + 3 ) ( x 2 + 3 x + 3 )
R = x = 1 ∑ ∞ [ ( x 2 − 3 x + 3 ) ( x 2 + 3 x + 3 ) x ]
= 6 1 x = 1 ∑ ∞ [ ( x 2 − 3 x + 3 ) 1 − ( x 2 + 3 x + 3 ) 1 ]
( A T e l e s c o p i c S e r i e s )
= 6 1 [ 1 + 1 + 3 1 ] = 1 8 7
∴ 1 8 − 7 = 1 1