2015 Countdown Problem #9: Maclaurin 2015

Calculus Level 3

Find the coefficient of the x 2015 x^{2015} term in the Maclaurin expansion of sin ⁡ ( x 2 ) ( 1 + 5 x 2 ) \frac{\sin{(x^2)}}{(1+5x^2)}

This problem is part of the set 2015 Countdown Problems .

None of the numbers listed. 50625 4060225 225

Wee Xian Bin by Wee Xian Bin , 25, Singapore

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1 solution

Wee Xian Bin
Dec 13, 2014

sin ⁡ ⁡ ( x 2 ) = x 2 − x 6 6 + x 10 120 − x 14 5040 + ⋯ \sin⁡(x^2 )=x^2-\frac{x^6}{6}+\frac{x^{10}}{120}-\frac{x^{14}}{5040}+⋯ ( 1 + 5 x 2 ) − 1 = 1 − 5 x 2 + 25 x 4 − … (1+5x^2 )^{-1}=1-5x^2+25x^4-… sin ⁡ ⁡ ( x 2 ) ( 1 + 5 x 2 ) = ( x 2 − x 6 6 + x 10 120 − x 14 5040 + ⋯ ) ( 1 − 5 x 2 + 25 x 4 − … ) \frac{\sin⁡(x^2 )}{(1+5x^2 )}=(x^2-\frac{x^6}{6}+\frac{x^{10}}{120}-\frac{x^{14}}{5040}+⋯)(1-5x^2+25x^4-…) Realise that there are no terms in the expansion with odd powers of x x . Hence coefficient of the x 2015 x^{2015} term = 0

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