If n is an in integer greater than 1, prove that

a–C_{1}(a–1)+C_{2}(a–2)
– . . . + (–1)^{n} C_{n} (a–n) = 0.

LHS = a –C_{1}(a–1) + c_{2} (a–2) –. . .+ (–1)^{n}
C_{n} (a–n).

= a [1–C_{1}+C_{2}
– . . . (–1)^{n} C_{n}] + [C_{1}–2C_{2}+3C_{3}–
. . .+(–1)^{n–1} nC_{n}].

= aP + Q . . . .(1).

P = C_{0} –C_{1}
+ C_{2} – . . . + (–1)^{n} C_{n} = 0 . . . (2).

Also,

C_{0} + C_{1}x
+ C_{2}x^{2} + . . . + C_{n}x^{n} = (1+x)^{n}.

Differentiating w.r.t. x,.

C_{1} + 2C_{2}x
+ 3C_{3}x^{2} + . . . . + nC_{n}x^{n-1} =
n(1+x)^{n–1}.

Putting x = –1

C_{1} –2C_{2}
+ 3C_{3} –. . . + (–1)^{n–1} nC_{n}= 0
. . . (3).

LHS = aP + Q = a´ 0 + 0 = 0. .

If and a_{k} = 1 for all k ≥n. Then show that

b_{n} = ^{2n+1}C_{n + 1} .

Put x - 3 = y, we
get . We have to find coefficient of
y^{n} in Left hand side as coefficient of y^{n} in the right
hand side is b_{n}. And we will start getting coefficient of y^{n}
only when r ≥ n in the left hand
side.

Coefficient of y^{n}
in (1 + y)^{n} + (1 + y)^{n + 1} + ... + (1 + y)^{2n} =
b_{n}.

Coefficient of y^{n}
in (1 + y)^{n}

&⇒ Coefficient of y^{n + 1} in = b_{n }

&⇒ ^{2n+1}C_{n+1} = b_{n}
.

If (1 + x)^{n} = show that

rth factor of is given by

t_{r} =

Now =
t_{1} . t_{2} . t_{3} …… t_{n}

=

=

=

Prove that = (–1)^{n–1 }nC_{n}

(1+x)^{2n }=
^{2n}C_{0} + ^{2n}C_{1}x+^{2n}C_{2}x^{2}
+ . . . + ^{2n}C_{2n}x^{2n} . . . (1).

Differentiating w.r.t. x, we get .

2n(1+x)^{2n–1} = C_{1} +2C_{2}x
+ 3C_{3}x^{2}+ . . . +2nC_{2n}x^{2n-1} .
. . (2).

Also =
C_{0} – C_{1} +C_{2}
. . . + C_{2n} . . . (3)

Where C_{r} = ^{2n}C_{r}

Multiplying (2)_{ }and (3), we get,
2n(1+x)^{2n–1}

= (C_{1}+2C_{2} x +3C_{3}x^{2}
+ . . . +2nC_{2n} x^{2n-1}) (C_{0 }–C_{1}+C_{2}. . . C_{2n} )

The coefficient of in RHS is

=

Also, the coefficient of in 2n(1+x)^{2n-1}

= coefficient of in 2n.

= coefficient of in
(1–x^{2})^{2n–1}(1–x)

= coefficient of x^{2n–2} in 2n (1–x^{2})^{2n–1}(1–x)

= coefficient of x^{2n-1} in 2n (1–x^{2})^{2n–1}
´ coefficient of x
in(1–x)

= coefficient of (x^{2})^{n–1}
in 2n (1–x^{2})^{2n–1} ´
coefficient of x in (1–x)

= 2n (-1)^{n–1 2n–1}C_{n–1}(-1)
= (-1)^{n} 2n

= (-1)^{n} n = (-1)^{n} n = - (- 1)^{n–1} nC_{n} ….
(6)

From (5) and (6), we get, = (-1)^{n–1 }nC_{n }

Given s_{n} = 1 + q +q^{2}+........+q^{n}.

S_{n} =

Prove that ^{(n+1)}C_{1}
+ ^{(n+1)}C_{2}s_{1} + ^{(n+1)}C_{3}s_{2}+........+^{(n+1)}C_{n+1}s_{n}
= 2^{n }S_{n}.

s_{n} is in geometric progression,
hence

Consider ^{(n+1)}C_{1}
+ ^{(n+1)}C_{2}s_{1} + ^{(n+1)}C_{3}s_{3}+........+^{(n+1)}C_{n+1}s_{n}.

=

Since