Question

plz help me with this questions

Answer 1

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$sinx + sin(2x) + sin(3x) + ......... + sin(nx) = \frac{sin (\frac{n + 1}{2})x.sin \frac{nx}{2} }{sin \frac{x}{2} } \\ \\ step \: 1st \: = > for \: n = 1 \\ \\ l.h.s. = sinx \\ \\r.h.s. = \frac{sin( \frac{1 + 1}{2} )x.sin \frac{x}{2} }{sin \frac{x}{2} } = sinx \\ \\ thus \: the \: result \: is \: true \: for \: x = 1 \\ \\ step \: 2nd = > \\ \\ let \: it \: is \: true \: for \: n = k \\ \\ therefor \\ \\ sin(x) + sin(2x) + sin(3x) + ............. + sin(kx) = \frac{sin (\frac{k + 1}{2})x.sin \frac{kx}{2} }{sin \frac{x}{2} } ......eq _{1} \\ \\ now \\ \\ in \: the \: 3rd \: step \: we \: will \: try \: to \: prove \: that \: it \: is \: true \: for \: = > \: n \: = (k + 1) \\ \\ step \: 3rd = > \\ \\ sin(x) + sin(2x) + sin(3x) + ....... + sin(kx) + sin(k + 1)x \: = \: \frac{sin (\frac{k + 1}{2} ).sin \frac{kx}{2} }{sin \frac{x}{2} } + 2sin( \frac{k + 1}{2} )x.cos( \frac{k + 1}{2} )x \\ \\ = \frac{sin( \frac{k + 1}{2}) x.}{sin \frac{x}{2} } (sin \frac{kx}{2} + 2cos (\frac{k + 1}{2}) x.sin \frac{x}{2} ) \\ \\ = \frac{sin( \frac{k + 1}{2} )x}{sin \frac{x}{2} } (sin \frac{kx}{2} + sin(( \frac{k + 1}{2} )x + \frac{x}{2} ) - sin(( \frac{k + 1}{2} )x - \frac{x}{2} ) \\ \\ we \: know \: that \\ = > 2cos \alpha .sin \beta = sin( \alpha + \beta ).sin( \alpha - \beta ) \\ \\ = \frac{sin( \frac{k + 1}{2} )x}{sin \frac{x}{2} } (sin \frac{kx}{2} + sin( \frac{k + 2}{2} )x - sin \frac{k}{2} ) \\ \\ = \frac{sin( \frac{k + 1}{2} )x}{ sin\frac{x}{2} } .sin \frac{(k + 1) + 1}{2} x$

● Thus it is true for "n = (k+1)"

therefore it will be prove for all values of

=> n belongs to N. ______"PROVED"

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