For the sake of
convenience the coefficients ^{n}C_{0} , ^{n}C_{1}
, . . . , ^{n}C_{r} , . . . ,^{n}C_{n} are
usually denoted by C_{0} , C_{1} , . . . , C_{r} , . .
. ,C_{n} respectively .

- C
_{0}+ C_{1}+ C_{2}+. . . . . + C_{n}= 2^{n}. - C
_{0}- C_{1}+ C_{2}-. . . . . C_{n}= 0. - C
_{0}+ C_{2}+ C_{4}+. . . . . = C_{1}+ C_{3}+ C_{5}+. . . . . = 2^{n-1}. ^{n}C_{r}+^{n}C_{r-1 }=^{n+1}C_{r}- r
^{n}C_{r}_{}=n^{n-1}C_{r-1} - .
- If
^{n}C_{x}=^{n}C_{y}, then either x = y or x + y = n.

So, ^{n}C_{r}
= ^{n}C_{n - r} =

(i) Differentiating (1 + x)^{n }=
C_{0} + C_{1}x + C_{2}x^{2} + . . . + C_{n}x^{n
}_{, }on both sides _{}we .

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

x = 1

&⇒ n2^{n-1} = C_{1} +
2C_{2} + 3C_{3} + . . . +nC_{n}.

x = -1

&⇒ 0 = C_{1} - 2C_{2} +
. . . +(-1)^{n-1 }nC_{n}.

Differentiating (1) again and again we will have different results.

(ii) Integrating (1 + x)^{n} ,
we have,

+ (where C is a constant )

Put x = 0, we get C = -

Therefore …(1)

Put x = 1 in (1) we get

Put x = -1 in (1) we get,

(i) Problems
Related to series of Binomial coefficients in which each term is a product of
an integer and a binomial coefficient i.e. in the form k ^{n}C_{r}
.

If (1+x)^{n} = then prove that C_{1}
+ 2C_{2} + 3C_{3}+ . . . + nC_{n}= n2^{n-1}

Method (i): By summation

r^{th} term
of the given series, t_{r} = r ^{n}C_{r}

&⇒ t_{r} = n ^{n-1}C_{r-1}

Sum of the series = =

= = n 2^{n-1} .

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

Differentiating (1) with respect to x

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

Putting x = 1 in (2)

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

(ii) Problems related to series of binomial coefficients in which each term is a binomial coefficient divided by an integer i.e. in the form of .

If (1+ x)^{n}
= , show that C_{0}
+

Method (i): rth term of the given series

t_{r} =

Required sum = =

= = .

Method (ii): By calculus

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

Integrating both the sides of (1) with respect to x between the limits 0 to x

&⇒ . . . . (2)

Substituting x = 1 in (2) , we get

.

(iii) Problem related to series of binomial coefficients in which each term is a product of two binomial coefficients.

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

C_{0}C_{1}
+ C_{1}C_{2} + . . . . + C_{n-1} C_{n} = .

We have

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

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

multiplying (1) and (2), we get

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

Equating the coefficient
of x^{n+1}, we get

C_{0}C_{1}
+C_{1}C_{2} + . . . + C_{n-1}C_{n} = ^{2n}C_{n+1}
= .

If ( 1+x)^{n}
= then prove that ^{}.

We have

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

and also ( x + 1)^{2n} = ^{2n}C_{0}
x^{2n }+ ^{2n}C_{1} x^{2n-1} +. . . + ^{2n}C_{2n}
….(2).

multiplying (1) and (2), we get (^{2n}C_{0}
– ^{2n}C_{1} x + ^{2n}C_{2} x^{2} + . .
. . + ( -1 )^{2n}. ^{2n}C_{2n}x^{2n} )( ^{2n}C_{0}
x^{2n }+ ^{2n}C_{1} x^{2n-1} +. . . + ^{2n}C_{2n})
= ( 1 – x^{2})^{2n} equating the coefficient of x^{2n},
we get .

^{}

(iv) Questions involving alternate binomial coefficients:

Evaluate C_{0} - C_{1} + C_{2} -
C_{3} +...+ (-1)^{n}C_{n}.

Here alternately +ve and - ve sign occur

This can be obtained by putting (-1) instead of 1 in place of x in

(1
+ x)^{n} = C_{0} + C_{1}x +...+ ^{n}C_{n}x^{n},
we get C_{0} - C_{1} +...+ (-1)^{n}C_{n} = 0.

Now
to obtain the sum C_{0} + C_{2} + C_{4} + ...we add (1
+ 1)^{n} and (1 - 1)^{n}.

Similarly, the cube roots of unity may be used to evaluate

C_{0}
+ C_{3} + C_{6} + ... OR C_{1} + C_{4} +... OR
C_{2} + C_{5} +...

put
x = 1, x = w, x = w^{2} in

(1
+ x)^{n} = C_{0} + C_{1}x +...+ C_{n}x^{n}
and add to get C_{0} + C_{3} + C_{6} +... the other two
may be obtained by suitably multiplying (1 + w)^{n} and .

(1
+ w^{2})^{n} by w and w^{2} respectively.

Prove that C_{0} + C_{4} + C_{8}
+ L =

Since (1+i)^{n} = C_{0} + C_{1}i
+ C_{2}i^{2} + C_{3}i^{3} + L +
C_{n}i^{n}

= (C_{0} - C_{2}
+ C_{4} -L ) + i(C_{1} - C_{3}
+ L)

But (1 + i)^{n} =

Hence = C_{0}
- C_{2}
+ C_{4} + L - - - (1)

Also 2^{n-1} = C_{0} + C_{2}
+ C_{4} + L - - - (2)

Adding (1) and (2) we have finally

C_{0} + C_{4} + C_{8} + L =

(v) Questions involving the greatest integer function:

These questions generally involve working with binomial expansions on surds.

If I is integral part of (2 + )^{n} and f is
fraction part of (2 + )^{n},
then prove that (I + f) (1 –f) = 1. Also prove that I is an odd Integer.

(2 + Ö3)^{n} = I + f where I is an
integer and 0 £ f < 1 show that I is odd and that (I + f) (1 - f)=1

Here note that (2 - Ö3)^{n} (2 +Ö3)^{n}
= (4 - 3)^{n} = 1

\ (2 + Ö3)^{n} (2 -Ö3)^{n}
= 1 it is thus required to prove that

(2 - Ö3)^{n} = 1 – f but, (2 - Ö3)^{n}
+ (2 +Ö3)^{n
}= [2^{n} - C_{1}.2^{n - 1}.Ö3 +
C_{2}2^{n - 2}..(Ö3)^{2 }- ...]

+ [2^{n} + C_{1}.2^{n - 1}.Ö3 +
C_{2}2^{n -2}..(Ö3)^{2} - ...] = 2[2^{n}
+ C_{2}.2^{n - 2}.3+C_{4}2^{n - 4}.3^{2}
+ ...].

= even integer

\ Now 0 < (2 - Ö3) < 1

\ 0 < (2 - Ö3)^{n} < 1

\ if (2 - Ö3)^{n} = f '

then I + f + f ' = Even

Now O £f < 1 and 0 < f ' < 1 . . . (1).

Also I + f + f ' = Even integer

\ f + f ' = Even integer . . . (2).

(1) and (2) imply that f + f ' = 1 (Q 0 < f + f ' < 2)

\ I is odd and f ' = 1 - f

&⇒ (I + f) (1 - f) =
1.