24thx soumya...
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Q1 was my dbt...i need just the ans for Q2..
Q1 r=0Σn nCr cosrx.sin(n-r)x =??
Q2 (1+nC0nC1)(1+nC2nC1).....(1+nCnnCn-1)=???
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2 Answers
1Q1 . First observe that ,
C1 cos x sin (n-1) x + Cn-1 sin x cos (n-1)x
=C1 { sin (nx - x + x) }------------------- [ as C1 = Cn-1 & sin a cos b + sin b cos a = sin (a+b) ]
=C1 sin nx
Similarly , C2 sin 2x cos(n-2)x + Cn-2 cos 2x sin (n-2)x = C2 sin nx
Now 2 x \sum_{r = 0}^{r = n} Cr cos rx sin (n-r) x
= ( C0 sin nx cos 0 + Cn sin 0 cos nx )
+ ( C1 sin (n-1)x cos x + Cn-1 sin x cos (n-1)x )
+ ( C2 sin (n-2)x cos 2x + Cn-2 sin 2x cos (n-2)x )
+ ...................... [ by pairing all like terms ]
= C0 sin nx + C1 sin nx + C2 sin nx ..........
= sin nx ( C0 + C1 + C2 + ......)
= 2n sin nx
So required summation = 2n-1 sin nx