Telescoping?

Geometry Level pending

T ( n ) = cos ⁡ 2 ( 3 0 ∘ − n ∘ ) − cos ⁡ ( 3 0 ∘ − n ∘ ) cos ⁡ ( 3 0 ∘ + n ∘ ) + cos ⁡ 2 ( 3 0 ∘ + n ∘ ) T(n) = \cos^2(30^\circ - n^\circ) - \cos(30^\circ -n^\circ) \cos(30^\circ + n^\circ) + \cos^2(30^\circ+n^\circ)

For T ( n ) T(n) as defined above, find the value of 4 ∑ n = 1 30 n T ( n ) \displaystyle 4\sum_{n=1}^{30} n T(n) .


The answer is 1395.

Elijah L by Elijah L , 49, Canada

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1 solution

Chew-Seong Cheong
Apr 25, 2020

Given that:

T ( n ) = cos ⁡ 2 ( 30 ° − n ° ) − cos ⁡ ( 30 ° − n ° ) cos ⁡ ( 30 ° + n ° ) + cos ⁡ 2 ( 30 ° + n ° ) = 1 2 ( 1 + cos ⁡ ( 60 ° − 2 n ° ) ) − 1 2 ( cos ⁡ ( 2 n ° ) + cos ⁡ 60 ° ) + 1 2 ( 1 + cos ⁡ ( 60 ° + 2 n ° ) ) = 1 2 + 1 4 cos ⁡ ( 2 n ° ) + 3 4 sin ⁡ ( 2 n ° ) − 1 2 cos ⁡ ( 2 n ° ) − 1 4 + 1 2 + 1 4 cos ⁡ ( 2 n ° ) − 3 4 sin ⁡ ( 2 n ° ) = 3 4 \begin{aligned} T(n) & = \cos^2(30\degree - n\degree) - \cos(30\degree - n\degree) \cos(30\degree + n\degree) + \cos^2(30\degree + n\degree) \\ & = \frac 12 (1+\cos(60\degree - 2n\degree)) - \frac 12 (\cos (2n\degree)+\cos 60\degree) + \frac 12 (1+\cos(60\degree + 2n\degree)) \\ & = \frac 12 + \cancel{\blue{\frac 14 \cos(2n\degree)}} + \cancel{\red{\frac {\sqrt 3}4 \sin (2n\degree)}} - \cancel{\blue{\frac 12 \cos(2n\degree)}} - \frac 14 + \frac 12 + \cancel{\blue{\frac 14 \cos(2n\degree)}} - \cancel{\red{\frac {\sqrt 3}4 \sin (2n\degree)}} \\ & = \frac 34 \end{aligned}

Therefore, 4 ∑ n = 1 30 n T ( n ) = 4 ∑ n = 1 30 3 n 4 = 3 ∑ n = 1 30 n = 3 ⋅ 30 ⋅ 31 2 = 1395 \displaystyle 4\sum_{n=1}^{30} n T(n) = 4\sum_{n=1}^{30} \frac {3n}4 = 3 \sum_{n=1}^{30} n = 3 \cdot \frac {30\cdot 31}2 = \boxed{1395} .

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Sir, earlier there used to be an option of seeing the LaTeX of the solution written , but now it seems that they have removed it, so is there anyway I can see the latex of your solution...because one can easily learn through seeing and we already know the output....so it used to be of great help.

I wanted to ask that can you bring that feature back since you are a moderator...

Vilakshan Gupta - 1 year, 1 month ago

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I think it costs too much for them to upgrade their server. I am not a staff I can't influence them

Chew-Seong Cheong - 1 year, 1 month ago

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