Prove that `sum_(k=0)^n ^nC_k sinKx.cos(n-K)x=2^(n-1)sin nx`

Options

A. `S_(5)(pi/2) =16`
B. `S_(7)((-pi)/2)=64`
C. `S_(50)(pi) = 0`
D. `S_(51)(-pi)=-2^(50`

Answer

a, b, c

Explanation

`therefore S_(n) (x) = sum_(k=0)^(n) ""^(n) C_(k) sin (kx) cos ((n-k)x)` ...(i)
Replace k by `n-k` in Eq, (i), then
` S_(n) (x) = sum_(k=0)^(n) ""^(n) C_(n-k) sin((n-k)x)cos (ks)`
or ` S_(n) (x) = sum_(k=0)^(n) ""^(n) C_(k) sin((n-k)x)cos (ks)` ...(ii)
On adding Eqs. (i) and (ii), we get `2S_(n) (x) = sum_(k=0)^(n) ""^(n) C_(k) cdot sin((nx)=2^(n) cdot sin(ks)`
`rArr S_(n)(x) = 2^(n-1) cdot sin (nx)`
`therefore S_(5)(pi/2) = 2^(4) cdot sin ((5pi)/2)=16,`
`S_(7)(-pi/2)=2^(6)cdot sin (-(7pi)/2) = 2^(6) xx - 1 xx-1 = 64`
`S_(50)(pi)=2^(49) cdot sin (50pi ) = 0`
and `S_(51)(-pi)=2^(50) cdot sin (-51pi ) = 0`