study material-mathematics-algebra-binomial theorem
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 sin2
m ?
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