"Series" feat. "Trigonometry"

Algebra Level 3

The first 2 2 terms of an arithmetic progression are 1 1 and cos ⁡ 2 x \cos^2 x respectively. The sum of the first 10 10 terms of the series, S 10 S_{10} ,can be written as a − b sin ⁡ 2 x a-b\sin^2 x where a a and b b are integers. Find a + b a+b .


The answer is 55.

Hoo Zhi Yee by Hoo Zhi Yee , 26, Malaysia

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1 solution

Hoo Zhi Yee
Apr 18, 2015

Let the common difference be d d . So, d = cos ⁡ 2 x − 1 d=\cos^2 x-1 .

S 10 = 10 2 [ 2 ( 1 ) + ( 10 − 1 ) ( cos ⁡ 2 x − 1 ) ] = 5 [ 2 + 9 ( cos ⁡ 2 − 1 ) ] = 10 + 45 ( cos ⁡ 2 x − 1 ) S_{10}=\frac{10}{2}[2(1)+(10-1)(\cos^2 x -1)]=5[2+9(\cos^2 -1)]=10+45(\cos^2 x-1)

But from sin ⁡ 2 x + cos ⁡ 2 x = 1 \sin^2 x+\cos^2 x=1 , we obtain

cos ⁡ 2 x − 1 = − sin ⁡ 2 x \cos^2 x -1= -\sin^2 x

Thus,

10 + 45 ( cos ⁡ 2 x − 1 ) = 10 − 45 sin ⁡ 2 x 10+45(\cos^2 x -1)=10-45\sin^2 x

Therefore, a + b = 10 + 45 = 55 a+b=10+45=55 .

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Same method..

Sagar Shah - 5 years, 2 months ago

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