f ( x ) = 7 x 2 0 1 7 + x 2 + x + 5
Let f ( x ) be a polynomial having roots x i for i = 1 to 2 0 1 7 .Then find the value of
1 0 6 1 ⎝ ⎛ n = 1 ∑ 2 0 1 7 i = 1 ∏ 2 0 1 7 ( x i + n ) − k = 1 ∑ 2 0 1 7 k 2 0 1 7 ⎠ ⎞
Details and Assumptions:
You can use calculator for lengthy calculations.
Write answer till 2 decimal places
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i did same and nice problem
Exactly the same way. Well done!!
Yes I had similar method.
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This problem is quite similar to this problem . Nice indeed! Also Congrats for having 400 followers!
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Thanks and congrats for yours 1404.And it is inspired from it after solving that problem I started looking for patterns in generalized thing and result is this problem.
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@Shivamani Patil – If thats so , then you must appreciate that problem by linking it with title "Inspiration" because if that problem was not generated , this problem by you also would not have generated, :) FYI: Its quite good to get inspired ;)
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@Nihar Mahajan – Ya I mentioned.Actually I posted this problem late night so forgot it :)
@Nihar Mahajan – By the way, I'll cross the 100 followers mark today. My 100 Follower Problem would be deadly. ":D
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@Satyajit Mohanty – " Curiosity at its best "
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@Nihar Mahajan – Yes his problems are good BTW.
@Satyajit Mohanty – On what topic you are going to make it?
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@Shivamani Patil – Don't know. Well, I don't make problems on my own. I just twist problems that I had solved in the past. Let's see! Well, I won't make it too deadly. People should enjoy it.
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@Satyajit Mohanty – That depends upon how you define "deadly" and "too deadly"! :)I would definitely.
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f ( x ) can be rewritten as: f ( x ) = 7 i = 1 ∏ 2 0 1 7 ( x − x i ) Now, the sum can be expressed as: S = 1 0 6 1 ( n = 1 ∑ 2 0 1 7 i = 1 ∏ 2 0 1 7 ( x i + n ) − k = 1 ∑ 2 0 1 7 k 2 0 1 7 ) S = 1 0 − 6 ( n = 1 ∑ 2 0 1 7 ( − 1 ) − 2 0 1 7 [ i = 1 ∏ 2 0 1 7 − ( x i + n ) ] − k = 1 ∑ 2 0 1 7 k 2 0 1 7 ) S = 1 0 − 6 ( − n = 1 ∑ 2 0 1 7 [ i = 1 ∏ 2 0 1 7 ( − n − x i ) ] − k = 1 ∑ 2 0 1 7 k 2 0 1 7 ) S = − 1 0 − 6 ( n = 1 ∑ 2 0 1 7 7 f ( − n ) + n = 1 ∑ 2 0 1 7 n 2 0 1 7 ) S = − 1 0 − 6 ( n = 1 ∑ 2 0 1 7 − n 2 0 1 7 + 7 1 n 2 − 7 1 n + 7 5 + n 2 0 1 7 ) S = 7 − 1 0 − 6 ( n = 1 ∑ 2 0 1 7 n 2 − n + 5 ) S = 7 − 1 0 − 6 ( 6 2 0 1 7 ⋅ 2 0 1 8 ⋅ 4 0 3 5 − 2 2 0 1 7 ⋅ 2 0 1 8 + 2 0 1 7 ⋅ 5 ) S = 8 4 − 2 0 1 7 ( 2 ⋅ 2 0 1 8 ⋅ 4 0 3 5 − 6 ⋅ 2 0 1 8 + 6 0 ) ⋅ 1 0 − 6 S = 8 4 − 2 0 1 7 ( 2 0 1 8 ⋅ 8 0 6 4 + 6 0 ) ⋅ 1 0 − 6 ≈ − 3 9 0 . 7 5