binooo

Q1 r=0Σn nCr cosrx.sin(n-r)x =??

Q2 (1+nC0nC1)(1+nC2nC1).....(1+nCnnCn-1)=???

2 Answers

1
Maths Musing ·

Q1 . 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

24
eureka123 ·

thx soumya...

@light
Q1 was my dbt...i need just the ans for Q2..

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