r = 0 ∑ n n C r sin ( r x ) cos ( ( n − r ) x ) = f ( n ) sin ( n x )
Let f ( n ) denote a function such that the equation above is an identity.
Find the value of f ( 1 5 ) .
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Nice solution! +1
Did you mean c o s ( ( n − r ) x ) in the first summation and c o s ( r x ) in the second summation?
@Tanishq Varshney Hi Tanishq! Try this please.
@Tanishq Varshney I deleted the problem and reposted it again since it had a small error the first time. Please do try it now. Many thanks!
I don't believe I actually calculated the LHS. Glad that i got the answer , btw Nice solution . : )
Let ∑ k = 0 n ( k n ) sin ( k x ) cos [ ( n − k ) x ] = G ( x ) . It is obvious that
G ( x ) = 2 ( n − 1 ) × sin ( n x ) ∀ n positive odd integer. Therefore f ( 1 5 ) = 2 1 4 = 1 6 3 8 4 .
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P = r = 0 ∑ n n C r s i n ( r x ) c o s ( n − r ) x
P = r = 0 ∑ n n C n − r s i n ( n − r ) x c o s ( r ) x
adding
2 P = ( n C 0 + n C 1 + n C 2 + n C 3 + . . . . + n C n ) s i n ( n x )
2 P = 2 n s i n ( n x )
f ( n ) = 2 n − 1
f ( 1 5 ) = 2 1 4 = 1 6 3 8 4