400 followers problem

Algebra Level 5

f ( x ) = 7 x 2017 + x 2 + x + 5 \large f\left( x \right) =7{ x }^{ 2017 }+{ x }^{ 2 }+x+5

Let f ( x ) f\left( x \right) be a polynomial having roots x i { x }_{ i } for i = 1 i=1 to 2017 2017 .Then find the value of

1 10 6 ( n = 1 2017 i = 1 2017 ( x i + n ) k = 1 2017 k 2017 ) \large \frac { 1 }{ { 10 }^{ 6 } } \left( \sum _{ n=1 }^{ 2017 }{ \prod _{ i=1 }^{ 2017 }{ \left( { x }_{ i }+n \right) } } -\sum _{ k=1 }^{ 2017 }{ { k }^{ 2017 } } \right)

Details and Assumptions:

  • You can use calculator for lengthy calculations.

  • Write answer till 2 decimal places

Inspiration .

This problem is original,


The answer is -390.75.

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1 solution

Daniel Turizo
Jul 17, 2015

f ( x ) f(x) can be rewritten as: f ( x ) = 7 i = 1 2017 ( x x i ) f(x) = 7\prod\limits_{i = 1}^{2017} {\left( {x - x_i } \right)} Now, the sum can be expressed as: S = 1 1 0 6 ( n = 1 2017 i = 1 2017 ( x i + n ) k = 1 2017 k 2017 ) S = \frac{1}{{10^6 }}\left( {\sum\limits_{n = 1}^{2017} {\prod\limits_{i = 1}^{2017} {\left( {x_i + n} \right) - \sum\limits_{k = 1}^{2017} {k^{2017} } } } } \right) S = 1 0 6 ( n = 1 2017 ( 1 ) 2017 [ i = 1 2017 ( x i + n ) ] k = 1 2017 k 2017 ) S = 10^{ - 6} \left( {\sum\limits_{n = 1}^{2017} {\left( { - 1} \right)^{ - 2017} \left[ {\prod\limits_{i = 1}^{2017} { - \left( {x_i + n} \right)} } \right] - \sum\limits_{k = 1}^{2017} {k^{2017} } } } \right) S = 1 0 6 ( n = 1 2017 [ i = 1 2017 ( n x i ) ] k = 1 2017 k 2017 ) S = 10^{ - 6} \left( { - \sum\limits_{n = 1}^{2017} {\left[ {\prod\limits_{i = 1}^{2017} {\left( { - n - x_i } \right)} } \right] - \sum\limits_{k = 1}^{2017} {k^{2017} } } } \right) S = 1 0 6 ( n = 1 2017 f ( n ) 7 + n = 1 2017 n 2017 ) S = - 10^{ - 6} \left( {\sum\limits_{n = 1}^{2017} {\frac{{f\left( { - n} \right)}}{7} + \sum\limits_{n = 1}^{2017} {n^{2017} } } } \right) S = 1 0 6 ( n = 1 2017 n 2017 + 1 7 n 2 1 7 n + 5 7 + n 2017 ) S = - 10^{ - 6} \left( {\sum\limits_{n = 1}^{2017} { - n^{2017} + \frac{1}{7}n^2 - \frac{1}{7}n + \frac{5}{7} + n^{2017} } } \right) S = 1 0 6 7 ( n = 1 2017 n 2 n + 5 ) S = \frac{{ - 10^{ - 6} }}{7}\left( {\sum\limits_{n = 1}^{2017} {n^2 - n + 5} } \right) S = 1 0 6 7 ( 2017 2018 4035 6 2017 2018 2 + 2017 5 ) S = \frac{{ - 10^{ - 6} }}{7}\left( {\frac{{2017 \cdot 2018 \cdot 4035}}{6} - \frac{{2017 \cdot 2018}}{2} + 2017 \cdot 5} \right) S = 2017 84 ( 2 2018 4035 6 2018 + 60 ) 1 0 6 S = \frac{{ - 2017}}{{84}}\left( {2 \cdot 2018 \cdot 4035 - 6 \cdot 2018 + 60} \right) \cdot 10^{ - 6} S = 2017 84 ( 2018 8064 + 60 ) 1 0 6 390.75 S = \frac{{ - 2017}}{{84}}\left( {2018 \cdot 8064 + 60} \right) \cdot 10^{ - 6} \approx \boxed{ - 390.75}

i did same and nice problem

Dev Sharma - 5 years, 5 months ago

Exactly the same way. Well done!!

Vishwak Srinivasan - 5 years, 11 months ago

Yes I had similar method.

shivamani patil - 5 years, 11 months ago

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This problem is quite similar to this problem . Nice indeed! Also Congrats for having 400 followers!

Nihar Mahajan - 5 years, 11 months ago

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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.

shivamani patil - 5 years, 11 months ago

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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 ;)

Nihar Mahajan - 5 years, 11 months ago

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@Nihar Mahajan Ya I mentioned.Actually I posted this problem late night so forgot it :)

shivamani patil - 5 years, 11 months ago

@Nihar Mahajan By the way, I'll cross the 100 followers mark today. My 100 Follower Problem would be deadly. ":D

Satyajit Mohanty - 5 years, 11 months ago

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@Satyajit Mohanty " Curiosity at its best "

Nihar Mahajan - 5 years, 11 months ago

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@Nihar Mahajan Yes his problems are good BTW.

shivamani patil - 5 years, 11 months ago

@Satyajit Mohanty On what topic you are going to make it?

shivamani patil - 5 years, 11 months ago

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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.

Satyajit Mohanty - 5 years, 11 months ago

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@Satyajit Mohanty That depends upon how you define "deadly" and "too deadly"! :)I would definitely.

shivamani patil - 5 years, 11 months ago

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