Sum till 2015, to welcome 2016
n = 1 ∑ 2 0 1 5 ( n + n + 1 ) ( 4 n + 4 n + 1 ) 1
If the above summation can be expressed as a 4 b − c where a , b , c are positive integers and b is free of fourth power, find the value of a + b + c .
The answer is 129.
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3 solutions
Ummm... 126 isnt exactly square-free... Maybe you should say it's fourth-power-free... XD
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Hello Nihar...although question solved....but there is a problem
1 2 6 = b is not square free as it is divisible by a perfect square 9
You should write b is not divisible by any perfect fourth power other than 1
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@Manuel Kahayon @Ravi Dwivedi Thanks for your attentiveness , I have updated the problem statement :)
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@Nihar Mahajan – Someone else feel this problem got rated too much?
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@A Former Brilliant Member – Well, I dont know any exact reason why a problem about telescoping series should be at level 5...
@A Former Brilliant Member – Thanks , I have set the level to Level 4 again.
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@Nihar Mahajan – Wait, you can do that?
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@Manuel Kahayon – Yes , I am a moderator :)
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@Nihar Mahajan – You can change levels at will?
@Nihar Mahajan – By the way what is a moderator And how brilliant selects them
@Nihar Mahajan – Why do the ratings oscillate?It's again level 5.
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@Rohit Udaiwal – Changed to Level 3. I want it to be 120 points.
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@Nihar Mahajan – This problem has gained twice the rating it should be! . the problem under any circumstances do not deserve to be level 5!
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@Prakhar Bindal – I have already edited the ratings three times before , but it is tending to level 5 again and again.
Rishabh, that's cool!
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That's why he is named Rishabh Cool :P
Thanks..... :)
Rationalize the denominator and it'll become a telescoping sequence.
= n = 1 ∑ 2 0 1 5 ( n + n + 1 ) ( 4 n + 4 n + 1 ) 1 = n = 1 ∑ 2 0 1 5 ( n + n + 1 ) ( n − n + 1 ) 4 n − 4 n + 1 = n = 1 ∑ 2 0 1 5 − 1 4 n − 4 n + 1 = n = 1 ∑ 2 0 1 5 4 n + 1 − 4 n = n = 2 ∑ 2 0 1 6 4 n − n = 1 ∑ 2 0 1 5 4 n = 4 2 0 1 6 − 4 1 = 2 4 1 2 6 − 1 ⇒ 2 + 1 2 6 + 1 = 1 2 9