A geometry problem by Archit Tripathi

Geometry Level 5

The value of

∑ r = 0 6 ( 6 r ) sin ⁡ ( r + 1 ) x \sum_{r=0}^6 {6\choose r} \sin(r + 1)x

is equal to ( 2 cos ⁡ ( x 2 ) ) p sin ⁡ ( q + 1 ) x (2\cos(\frac{x}{2}))^{p}\sin(q + 1)x .

Find the value of 25 ( p + q ) 3 \frac{25(p + q)}{3} .


The answer is 75.

Archit Tripathi by Archit Tripathi , 23, India

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1 solution

Archit Tripathi
Dec 10, 2016

Imaginary part of e i x ( 1 + e i x ) n e^{ix}(1 + e^{ix})^{n}

e i x ( 2 c o s ( x 2 ) ) n e i n x / 2 e^{ix}(2cos(\frac{x}{2}))^{n}e^{inx/2}

= ( 2 c o s ( x 2 ) ) n [ c o s ( n 2 + 1 ) x + i s i n ( n 2 + 1 ) x ] (2cos(\frac{x}{2}))^{n}[cos(\frac{n}{2} + 1)x + i sin(\frac{n}{2} + 1)x]

= ( 2 c o s ( x 2 ) ) n s i n ( n 2 + 1 ) x (2cos(\frac{x}{2}))^{n}sin(\frac{n}{2} + 1)x

Now, put n = 6 n = 6 , we get p = 6 p = 6 and q = 3 q = 3 which gives 25 ( 9 3 ) = 75 25(\frac{9}{3}) = \boxed{75}

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