21yes/// ans is 2^n-1 sin nx
this is called reversing technique..
3 Answers
62First Observe...
S=\sum_{k=0}^n{C(n,k)\sin(kx)\cos(n-k)x}=\sum_{k=0}^n{C(n,k)\sin(n-k)x\cos(k)x} (Using k=n-k)
Then see that this is the Sum of these two is
2S=\sum_{k=0}^n{C(n,k)\sin(kx)\cos(n-k)x}+\sum_{k=0}^n{C(n,k)\sin(n-k)x\cos(k)x}
2S=\sum_{k=0}^n{C(n,k)[\sin(kx)\cos(n-k)x}+\sin(n-k)x\cos(k)x}]=\sum_{k=0}^n{C(n,k)\sin(nx)=\sin(nx)\sum_{k=0}^n{C(n,k)=2^n\sin(nx)
71can you please tell me where from did you get this question:-).