Binomial sufficiency

Probability Level 4

∑ k = 0 1008 ( 2 k + 1 ) 2 ( 2017 2 k + 1 ) = a ( a + 1 ) 2 b \large\ \sum _{ { k=0 } }^{1008} { { \left( 2k+1 \right) } } ^{ 2 }{2017\choose{2k+1}}= a(a + 1)2^b

Above sum can be represented for positive integers a , b a, b .

Find a + b a + b .


The answer is 4031.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

Alan Yan
Jan 14, 2018

Let f ( x ) = ( 1 + x ) 2017 + ( x − 1 ) 2017 2 . f(x) = \frac{(1 + x)^{2017} + (x - 1)^{2017}}{2}. Define g ( x ) = f ′ ( x ) + x f ′ ′ ( x ) . g(x) = f'(x) + xf''(x). Our answer is just g ( 1 ) g(1) which gives 2017 ⋅ 2016 ⋅ 2 2015 2017 \cdot 2016 \cdot 2^{2015} . Hence, the answer is 4031 \boxed{4031} .

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