Given-f-x-1-x-x-2-x-3-x-2015-x-2016-2-Find-the-sum-of-all-odd-coeffisiens-Ex-f-x-x-2-x-1-2-1x-4-2x-3-3x-2-2x-1-The-sum-of-odd-coeffisien-is-1-

Question Number 26087 by Joel578 last updated on 19/Dec/17

Given

f (x) = {(1 - x + x^{2} - x^{3} + \dots - x^{2015} + x^{2016})}^{2}

Find the sum of all odd coeffisiens!

Ex . f (x) = {(x^{2} + x + 1)}^{2} = 1 x^{4} + 2 x^{3} + 3 x^{2} + 2 x + 1

The sum of odd coeffisien is 1 + 3 = 4

Commented by moxhix last updated on 19/Dec/17

f (1) = (sum of odd coeffisiens) + (sum of even coeffisiens)

f (- 1) = - (sum of odd coeffisiens) + (sum of even coeffisiens)

∴ (sum of odd coeffisiens) = \frac{1}{2} (f (1) - f (- 1))

E x . f (x) = {(x^{2} + x + 1)}^{2}

f (1) = 9, f (- 1) = 1

∴ (sum of odd coeffisiens) = \frac{1}{2} (9 - 1) = 4

L e t f (x) = {(1 - x^{1} + x^{2} - x^{3} + \dots - x^{2015} + x^{2016})}^{2}

f (1) = 1, f (- 1) = 2017^{2}

∴ (sum of odd coeffisiens) = \frac{1 - 2017^{2}}{2}

= - \frac{2016 \times 2018}{2} = - 2016 \times 1009

= - 2034144

Commented by moxhix last updated on 19/Dec/17

I^{'} m sorry, I misundertood \dots .

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