Binomial Theorem - 8

Also show that

nn-1 > (n + 1)n ; 3 ,

Ans . We get an =

= 1 + n + …………………

= 1 + 1 + + …………..

= 2 + Positive quantity

an > 2 . ……………(1)

Alos, an < 1 + 1 +

< 1 + 1 +

< 1 +

< 1 +

< 3 ……………….(2)

From (1) and (2) we get

2 < an < 3

So, an < 3

Or < 3

or < n ( n 3 )

or……….img

So, (1 + )n < n4. n

or (n + 1)n < nn+1

Dumb Question

Why img……….

img…………

Ans We are dividing one by a smaller no because all natural no.s except 2 are bigger than 2 only

So,

as

3 > 2 so,

For other factor also

So, 1 + 1 + is less than 1 + 1 +

Hard

(1) Prove that (-1 ) r-1 3n C2r-1 = 0 where k=

is an even positive inteqer.

Ans Given n is an even positive inteqer.

Let n = 2m R = 3m

L.H.S. (-3)r-1 3nC 2r-1 = (-3) r-1 6m C2r-1

= 6m C1 - 3 6m C3 + 32 6m C5-……..+ (-3)3m-16m C6m -1 ……….(1)

(1 + i ) 6m = 6m Co + 6m C1 (i ) + 6m C2 (i )2+ ………+ 6mC6m-1(i ) 6m-1+ 6m C6m(i )6mimg …………….

or 26m =6m Co + 6m C1 (i ) + 6m C2 (i )2+ ………+ 6mC6m-1(i ) 6m-1+ 6m C6m(i )6m

or 26m( cos 2 + i sin 2 m )

=( 6mCo - 3 6m C2 + 32 6m - …..+ (-3) 3m 6m C6m) + i ( 6mCo - 3 6m C2 + 32 6m - …..+ (-3) 3m 6m C6m)

Companing imaqinary part on both sides we get .

( 6m C1 - 3 6m C3 +32 6mC5………..+ (-3) 3m-1 6mC6m-1 ) = 0

Or ( 6m C1 - 3 6m C3 +32 6mC5………..+ (-3) 3m-1 6mC6m-1 ) = 0

=> (-3) r-1 6m C2r-2 = 0

Or (-33) r-1 3n C2r-1 = 0 (where n = 2m)

Dumb Question: How did ( cos + i sin ) 6m be came cos 2 m + i sin2m ?

inmg……..

Ans We used Eulers Rule which is (( cos + sin ) m = cosm +sin

So, ( cos + i sin ) 6m = cos 6m+ i sin 6m)

= cos 2m + i sin 2m


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