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1. Binomial Theorem
Binomial Theorem
INTRODUCTION
Binomial is made of “Bi” which means “two”
and “nomial” which means an expression with
numbers or variables.
Binomial Expression
eg : x + y,
1 yz x
Note :
x + (y + z) is also a binomial expression
where ‘x’ is first term and ‘(y + z)’ is second
term. Similarly, (x + y) + (z + w) is a binomial
expression where (y +z) is first term and
(z + w) is second term.
Binomial Theorem
(x + y) 2 = x 2
(x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3
(x + y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4
When index is very high then use binomial
theorem as below
(x + y) n = n C 0 x n y 0
n C 1 x n– y 1
n C 2 x n– y 2
- … + nC r xn–ryr+ … + nC n xn–nyn
Where n∈N, x and y ∈ set of complex
numbers and nC 0 , nC 1 , nC 2 … are called binomial
coefficients normally written as C 0 , C 1 , C 2 …
(n hidden)
Proof of Binomial Expansion
(x + y)n^ = (x + y) (x + y)…(x + y) {n brackets}
Each term in the expansion is formed by taking
one letter from each bracket and multiplying
them together. For example, choose x from
all the bracket and multiplying them, xn^ is
obtained. This can be done in 1 (= nC 0 ) way.
Now, choose x from (n – 1) brackets and y
from remaining one bracket and multiplying
them, x n– y 1 is obtained. But number of ways
of choosing (n – 1) brackets out of ‘n’ is nC n– = nC 1 ways.
Definition
An algebraic expression which contains two dissimilar terms is
called binomial expression.
Definition
The formula by which any
positive integral power of a
binomial expression can be
expanded in the form of a series
is known as Binomial Theorem.
Point to Remember!!!
(i) The number of terms in the
expansion is (n + 1)
i.e. one more than the index
(n)
(ii) Tr+1 = n Crx n–r y r is called the
general term of the binomial
expansion.
(iii) Sum of powers of each term
in (x + y)n^ is n.
Binomial Theorem 2.
Sum of all binomial coefficients in expansion of (x + y) n
C 0 + C 1 + C 2 +…+ Cn = 2 n …(1)
Proof :
Put x = y = 1 in expansion of (x + y)n
(x + y) n = n C 0 x n y 0
n C 1 x n– y 1
n C 2 x n– y 2 +…+ n Cnx n–n y n
(1 + 1)n^ = nC 0 1 n 10 + nC 1 1 n-1^11 +…+ nC n 10 1 n
2 n = n C 0 + n C 1 +…+ n Cn.
Note :
(x + 2y)^2 = 2 C 0 x2–0(2y)^0 + 2 C 1 x2–1(2y)^1 + 2 C 2 x2–2(2y)^2
= 1x 2
y Here, 2 C 0
, 2 C
1
, 2 C
2 are binomial coefficients and 1, 4, 4
are coefficients.
y To find sum of binomial coefficients in expansion
of (x + 2y) 2 , use formula (1) and to find sum of all
coefficients in the expansion of (x+2y) 2 put x = y = 1,
i.e. [1 + 2(1)]^2 = 9
Hence n C 1 × 1 C 1 = n C 1 = n C 1 terms of the form
x n– y 1 are generated. Continuing in this way, n Cr
terms of the form x n–r y r are generated & the
process is done till all possible combinations
of powers of ‘x’ and ‘y’ are obtained. Adding
all (x + y) n = x n
n C 1 x n– y 1 +…+ n Crx n–r y r +…+ y n
Highlights
n n n (^) n n r r r r 1 r 0 r 0
x y C x y T
−
= =
∑ ∑
Where Tr+1 is general term
Also, ( ) ( )
n n n (^) n n r r r r 0
x y y x C y x
−
=
∑
Point to Remember!!!
Number of distinct terms in the
expansion of (x 1 +x 2 +…+x r )n^ is n+r–1C r–
. Which is also equal to
number of ways of distributing
‘n’ identical coins among ‘r’
persons.
Q.
Find number of distinct terms in the expansion of
(a) (x + y + z)^2 (b) (x + y + z)^8
Sol.
(a) n = 2, r = 3 n+r– Cr–1 = 3+2– C3–1 = 4 C 2 = 6
(b) n = 8 r = 3 n+r–1C r–
= 8+3–1C
3–
= 10 C
2
Binomial Theorem 4.
Q.
**Find coefficient of x 6 in (1 + 3x + 3x 2
Sol.
Given expansion is [(1 + x) 3 ] 15 = (1 + x) 45
∴ coefficient of x^6 in (1 + x)^45 is 45 C 6
Q.
Show that coefficient of x 5 in the expansion of (1 + x 2 ) 5
. (1 + x) 4 is 60.
Sol.
5 C 0 + 5 C 1 x 2
5 C 2 x 4
5 C 3 x 6
5 C 4 x 8
5 C 5 x 10 ) × ( 4 C 0 + 4 C 1 x + 4 C 2 x 2
4 C 3 x 3
4 C 4 x 4 )
Required coefficient is (^5) C 1
4 C
3
+ 5 C
2
4 C
1
Q.
Let (^) ( ) ( )
n 4 (^2) n 2 k k k 0
1 x. 1 x a x
=
. If a 1 , a 2 and a 3 are in A.P. find n.
Sol.
(1 + 2x^2 + x^4 ) (1 + nC 1 x + nC 2 x^2 +…) =
a 0 x^0 + a 1 x^1 + a 2 x^2 +…
comparing coefficients of like powers of x both side
a 1 = n C 1 , a 2 = 2 + n C 2 , a 3 = n C 3 + 2( n C 1 )
2a 2 = a 1 + a 3
( ) n ( ) ( )
1
n n 1 n n 1 n 2 2 2 C 2n 2 6
6[4 + n(n – 1)] = 18n + n^3 – 3n^2 + 2n
24 + 6n^2 – 6n = 18n + n^3 – 3n^2 + 2n
n 3
(n – 2) (n – 3) (n – 4) = 0 ⇒ n = 2, 3, 4
Q.
The sum of coefficients of integral powers of x in the binomial expansion of
( )
50 1 − 2 x is
(A) (^) ( )
(D) (^) ( )
Ans.
(B)
Sol. (^ )^ (^ )^ (^ )^ (^ )
50 1 2 3 50 50 50 50 1 − 2 x = C 0 − C 1 2 x + C 2 2 x − C 3 2 x
( ) ( ) ( )
4 5 50 50 50 50 4 5 50
50
- C 2 x − C 2 x + … + ( 1)− C 2 x …(i)
5. Binomial Theorem
Required sum = 50 C 0 + 50 C 2 2 2
50 C 42 4 +…+ 50 C 502 50
50 50 50 1 50 2 50 3 50 50 (1 + 2 x ) = C 0 + C (2 1 x ) + C (2 2 x ) + C (2 3 x ) + … + C 50 (2 x )
…(ii)
(i) + (ii) gives
( )^ ( )^ ( )^ ( )^ ( )
50 50 2 4 50 50 50 50 50 0 2 4 50 1 2 x 1 2 x 2 C C 2 x C 2 x ... C 2 x
put x = 1 both side
50 50 1 2 1 2
2
− + +
50 C 0 + 50 C 22 2
50 C 42 4 +…+ 50 C 502 50
On solving L.H.S. is (^) ( )
Q.
Sum of all coefficients in (1 + x – 3x^2 )^2163
Sol.
Put x = 1
(1 + 1 – 3) 2163 = – 1
Q.
The sum of the coefficients in the expansion of (a + 2b + c)^10 is
(A) 4^10 (B) 3^10 (C) 2^10 (D) 10^4
Ans.
(A)
Sol.
Put a = b = c = 1
(1 + 2(1) + 1)^10 = 4^10
Q.
Find the fourth term in the expansion of
7 y 2x 2
−
Sol.
3 7 7 3 4 3 1 3
y T T C (2x) 2
−
3 7 6 5 4 y 16x ( 1) 6 8
× ×
= × −
= – 70x^4 y^3
7. Binomial Theorem
Q.
The coefficient of (3r) th term and coefficient of (r + 2) th term in the expansion
of (1 + x) 2n are equal (r > 1, n > 2 and positive integer) then
(A)
n r 2
= (B)
n r 3
= (C)
n 1 r 2
= (D)
n 1 r 2
−
Ans.
(A)
Sol.
2n C3r–1 = 2n Cr+
n n x y 3r 1 r 1 or 2n 3r 1 r 1 C^ C
2r 2 or 2n 4r x y n or
r 1 or n 2r x y
Q.
If in the expansion of (1 + y)n, then coefficient of 5th, 6th^ and 7th^ terms are in
A.P., then ‘n’ is equal to
(A) 7, 11 (B) 7, 14 (C) 8, 16 (D) None
Ans.
(B)
Sol.
2 nC 5 = nC 4
2 n! (^) n! n!
5! n 5! 4! n 4! 6! n 6!
12(n – 4) = 30 + (n – 5) (n – 4)
n 2
Q.
If ( ) ( )
2n 2n r r r r r 0 r 0
a x 2 b x 3
= =
∑ ∑
and a k = 1 ∀ k ≥ n, then show that b n = 2n+1C n+
Sol.
a 0 (x – 2)^0 + a 1 (x – 2)^1 + a 2 (x – 2)^2 +…+ a n– (x – 2)n–1^ + a n (x – 2)n^ + a n+ (x –2)n+ 1+…+
a2n(x – 2) 2n =
b 0 (x – 3) 0
- b 1 (x – 3) 1
- b 2 (x – 3) 2 +…+ b2n(x – 3) 2n
a 0
- a 1 (x – 2)^1 + a 2 (x – 2)^2 +…+ a n– (x – 2)n–1^ + (x –2)n^ + (x – 2)n+1^ +…+ (x – 2)2n
= b 0
- b 1 (x – 3)^1 + b 2 (x–3)^2 +…+ b n (x – 3)n^ + b n+ (x – 3)n+1^ +…+ b 2n (x – 3)2n
(M 1)
Comparing coefficient of x n from both sides- n C 0 + n+ C 1 + n+ C 2 +…+ 2n Cn = bn
b n = (n+1C n+
- n+1C n ) + n+2C n +…+ 2nC n = (n+2C n+
- n+2C n )+…+ 2nC n
Binomial Theorem 8.
[
n Cr + n Cr+1 = n+ Cr+1]
= ( (n+3) Cn+1 + n+ Cn) +…+ 2n Cn
2n+ Cn+
(M 2)
(x + 1)n^ + (x + 1)n+1^ + (x + 1)n+2^ +…+ (x + 1)2n^ = S(say)
To find coefficient of x n from above series.
A = (x + 1) n R = (x + 1)
n 1 n R 1 n^ x^1 S=A x 1 R (^1) x 1 1
− +^ −
=^ +
− ^
+^ −
2n 1 n 2n 1 n x 1 x 1 x 1 x 1 S x x x
∴ coefficient of x n in S = 2n+ Cn+1 (coefficient of x n+ in (1 + x) 2n+ )
∴ b n = 2n+1C n+
Term Independent of x
Step-I: Write general term of the expansion.
Step-II: Equate the exponent of x to zero
Q.
Find the term independent of x in
6 2 1 2x 3x
−
Sol. (^ )^
r 6 r 6 2 r 1 r
T C 2x 3x
−
r 6 6 r 12 2r r (^) r
C 2 x (^3) x
− − ^
r 6 6 r 12 3r r
C 2 x 3
− ^ −
For term independent of x, 12 – 3r = 0
r = 4
Hence T4+1 = T 5 or 5 th term is independent of x
4 6 6 4 0 5 4
T C 2 x 3
− ^
= × × =
Binomial Theorem 10.
Q.
Find the coefficient of x 15 in the expansion of (x – x 2 ) 10 .
Sol.
[x(1 – x)] 10 = x 10 (1 – x) 10
∴ coeff. of x^15 in x^10 (1 – x)^10 =
coeff. of x^5 in (1 – x)^10 = (-1)^5 10 C 5 = – 252
Q.
If {P} denotes the fractional part of the number P, then
200 3
8
, is equal to
(A)
5
8
(B)
1
8
(C)
7
8
(D)
3
8
Ans.
(B)
Sol.
200 = (1 + 8) 100 = 1 + 100 C 1 8 1
100 C 2 8 2 +…
= 1 + 8k, k ∈ integer
200 3 1 8k 1 1 1 k 8 8 8 8 8
=^ =^ +^ =^ =
Q.
Value of (^) ( ) ( )
7 7 1 + 2 + 1 − 2
(A) 402 (B) 458 (C) 478 (D) 498
Ans.
(C)
Sol. (^ )^ (^ )^ (^ )^ (^ )
7 1 2 3
1 2 3
1 + 2 = 1 + 7C 2 + 7C 2 + 7C 2 + … +
( ) ( ) ( ) ( )
7 1 2 3 1 − 2 = 1 − 7C 1 2 + 7C 2 2 − 7C 3 2 +...
( ) ( ) (^ )
7 7 7 7 2 7 3 1 2 1 2 2 1 C 2 2 C 2 4 C 2 6
= 2[1 + 42 + 140 + 56]
= 478
Q.
Find value of (^) ( ) ( )
5 5 3 + 3 − 3 − 3
Sol. (^ )^ (^ )
5 5 5 3 1 3 1 3
+^ −^ −
11. Binomial Theorem
( ) ( ) ( ) ( )
5 1 3 5 5 5 5 3 2 C 1 3 C 3 3 C 5 3
= 54[5 + 30 + 9]
= 54 × 44 = 2376
Q.
In the expansion of (1 + x) 10 , coefficients of (4r + 5) th term = coefficient of
(2r + 1)th^ term. Find r
Sol.
10 C4r + 4 = 10 C2r
4r + 4 = 2r or 10 = (4r + 4) + 2r ⇒ r = 1
2r = – 4 (not possible)
[nC x = nC y ⇒ x + y = n or x = y]
Q.
Let α > 0, β > 0 be such that α 3 + β 2 = 4. If the maximum value of the term
independent of x in the binomial expansion of
10 1 1
x 9 x^6
− (^) α + β
is 10k, then k is
equal to
(A) 176 (B) 336 (C) 352 (D) 84
Ans.
(B)
Sol.
10 r r 1 1 10 1
9 6 r r T C x x
− − (^) α (^) β
=
For term independent of x,
10 r r 0 9 6
− = ⇒ r = 4
∴ T
5
= 10 C
4 α^6 β^4
( )
3 2 1 (^3 2 )
2
α + β ≥ α β
2 (^43 )
2
≥ α β
⇒ (α^3 β^2 ) max
∴ (T
5
max = 10 k 10 C 4 (4) 2 = 10 k
k = 336
13. Binomial Theorem
Q.^20
1 3 1 4
1 4
6
. Find number of rational terms.
Sol.
2 20
r
r 1
0 r 1 1 3 4 r
T C 4 6
− −
For rational terms, ‘r’ should be multiple of 4
∴ Possible values of r is 0, 4, 8, 12, 16, 20
but at r = 8 and 20 only
20 r
3
is integer
∴ r = 8,
Hence, number of rational terms are 2
Q.
In the expansion of y = 1 + (1 + x) + (1 + x)^2 +…+ (1 + x)^19 , if the coefficient of xp
is greatest. Find p
Sol.
20 20 1 x 1 1 x 1 y 1 x 1 x
∴ coefficient of xp^ in y is 20 C p+ (^20) C p+ is greatest when p + 1 = 10 ⇒ p = 9.
Q.
Show that
2 2 2 2 2 2n C 0 + C 1 + C 2 + C 3 + ... + Cn = Cn
(M 1) By method of selection.
Consider n boys and n girls in a class. Total number of ways to select n
students out of 2n is 2nC n
. Alternatively, the same selection can be done if out
of n boys, zero boys and out of n girls, n girls are selected or out of n boys, 1
boy and out of n girls, (n–1) girls are selected and so on…
Hence 2nC n = nC 0
nC n
nC n–
nC n–
= nC 0
nC 0
nC 1
nC 2
∴ 2nC n
2 2 2 0 1 2 C + C + C +...
(M 2) Multiplication of two series
(1 + x)n^ = C 0
+ C
1 x + C 2 x^2 + …
Replace x by
x
Binomial Theorem 14.
n 1 2 (^0 )
1 C^ C
1 C ...
x x (^) x
+^ =^ +^ +^ +
( )
n n (^2 1 ) 0 1 2 0 2
1 C^ C
(1 x) 1 C C x C x C x x x
^ ^
2n
1 2 1 2 2 1 2 n 0 0 2 1 0 2 0 2 2
1 x (^) C C C C C C C C ... C x C ... C x C ... x x^ x x^ x^ x x
Taking coefficient of x 0 from both side
coefficient of x n in (1 + x) 2n =
2 2 2 2 C 0 + C 1 + C 2 + ... +Cn
2n Cn =
2 2 2 2 0 1 2 n C + C + C + ... +C
Q.
Find S where ( )
2 2 2 2 0 1 2 n S = 1.C + 2.C + 3.C + ... + n +1 C
Sol.
2 2 2 2 S = 1.C 0 + 2.C 1 + 3.C 2 + ... + n +1 Cn
2 2 2 +S = n + 1 Cn + n Cn (^) − 1 + ... + 1. C 0
2 2 2 0 1 n 2 S = n + 2 C + n + 2 C +... + n+2 C
( ) 2n
n
n 2 C 2
S
= ×
Q.
Prove that (^72 C 36
- 1) is divisible by 73.
Sol.
72 C 36 = 73 C 36 – 72 C 35 ( n Cr+1 = n+ Cr+1 – n Cr)
= 73 C 36
– (^73 C
35
– 72 C
34
= 73 C
36
– 73 C
35
+ (^73 C
34
– 72 C
33
72 C
36
–1 = 73 C
36
– 73 C
35
+ 73 C
34
– 73 C
33
+ 73 C
32
– … – 73 C
1
+ 72 C
0
– 1]
= [
73 C 36 – 73 C 35 + 73 C 34 – 73 C 33 + 73 C 32 – … – 73 C 1 ]
Each of the term contains a factor 73.
Hence, (^72 C 36– ) is divisible by 73.
Binomial Theorem 16.
10 10 20 10 20 30 10 2 20 30 10 10 r r 10 r 10 10 r 0 r 0
C C (^) − C C ( C ) C C
= =
(^) − (^) − + ^ ^
20 C 10 30 C 10 – 30 C 10 20 C 10 + 30 C 10 – 20 C 10 (^30) C 10
– 20 C
10
= C
10
– B
10
Q.
The value of r for which 20 Cr 20 C 0 + 20 Cr– 20 C 1 + 20 Cr– 20 C 2 +…+ 20 C 0 20 Cr is
maximum is
(A) 15 (B) 20 (C) 11 (D) 10
Ans.
(B)
Sol.
Out of 20 boys and 20 girls, total ‘r’ students are selected. This is stated by the
given series.
Hence it equals (20+20)C r
= 40 C
r
(^40) C r is maximum at r =
40 20 2
=
Q.
Show that C 0 + 2C 1 + 3C 2 + 4C 3 +…+ (n + 1) Cn = (n + 2). 2 n–
Sol.
(M 1)
S = C
0
+ 2C
1
+ 3C
2
+ 4C
3 +…+ (n + 1)C n
- S = (n + 1) Cn + nCn–1 +…+ 2C 1 + C 0
2S = (n + 2) [C 0 + C 1 +…+ Cn]
S =
n 2
2
[ n ]= (n + 2) 2 n–
(M 2)
Consider
(1 + x) n = C 0 + C 1 x + C 2 x 2 +…+ Cnx n
d n d 1 x x dx dx
[C 0 x + C 1 x 2
(1 + x)n^ + nx(1 + x)n–1^ = C 0
+ 2C
1 x + 3C 2 x^2 + … + (n + 1)C n xn
Put x = 1
2 n^ + n2n–1^ = C 0
+ 2C
1
+ 3C
2
- … + (n + 1)C n ∴ C 0 + 2C 1 + 3C 2 + … + (n + 1)Cn = (n + 2) 2 n–
17. Binomial Theorem
Q.
Show that S = 1.C 1 + 2.C 2 + 3.C 3 +…+ n.Cn = n. n–
Sol.
(M 1)
Consider
(1 + x) n = C 0 + C 1 x + C 2 x 2 +…+ Cnx n
Differentiate both sides with respect to x n(1 + x)n–1^ = C 1
+ 2C
2 x +…+ n.C n xn–
Put x = 1 n(2)n–1^ = C 1
+ 2C
2
+ 3C
3 +…+ n.C n (M 2)
Tr = r. n Cr
= n. n– Cr–1 ( )
n n 1 r r 1
n C C r
− −
n n n 1 n 1 r 1 r 1 r 1 r 1
S n. C C
n
− − − − = =
= =
= n.( n– )
Q.
Which is larger: (99^50 + 100^50 ) or (101)^50
Sol.
50
50 ↔ (1 + 100) 50
10050 ↔ (C
0
+ C
1 x + C 2 x^2 +…) – (C 0
– C
1 x + C 2 x^2 …) (x = 100)
10050 ↔ 2[^50 C 1
100 + 50 C
3
1003 +…+ 50 C
49
10049 ]
50 ↔ 2[ 50 C 1 100 + 50 C 3100 3 +…] + 100 50
Hence right hand side is bigger than left hand side
⇒ 99 50
Q.
**Show that : 2n–2C n–
4n
n + 1
, n ∈ N n > 2
Sol.
2n– Cn–2 ≥ 1, 2. 2n– Cn–1 ≥ 2, 2n– Cn ≥ 1
Adding, LHS ≥ 4
Now, RHS =
4n 4 4 4 4 4 n 1 n 1
⇒ RHS < LHS.
19. Binomial Theorem
n n 1 r 1 r 1 r 0
C 2
n 1
=
∑
n n 1 r 1 r 1 r 0
C 2
n 1
=
∑
n 1 1 n 1 2 n 1 3 n 1 n 1 1 2 3 n 1
1 C 2 C 2 C 2 ... C 2 1
n 1
+ ^
1 n^1 1 2 1 n 1
+
n 1 3 1
n 1
−
Q.
Prove :
n 1 C 0 C 1 C 2 C 3 Cn (^1) n. ... 2 3 4 5 n (^2) n 1 n 2
Sol.
(1 + x) n = C 0 + C 1 x + C 2 x 2
( ) ( )
1 1 n (^2 ) 0 1 2 0 0
x 1 + x dx = C x + C x + C x +... dx ∫ ∫
Put (1 + x) = t dx = dt
2 1 2 3 4 n 0 1 2 1 0
x x x t 1 t dt C C C ... 2 3 4
∫
2 2 n 1 n 0 1 2 n
1 1
C C C C
t dt t dt ... 2 3 4 n 2
∫ ∫
2 2 n 2 n 1 0 1 2 n
1 1
t t C^ C^ C^ C ... n 2 n 1 2 3 4 n 2
+^ +
−^ =^ +^ +^ +^ +
+^ +^ +
( )
(^1) n 2 1 n 1 C 0 C 1 C 2 2 1 2 1 ... n 2 n 1 2 3 4
+^ +
+ ^ +
C 0 C 1 C 2 n 1 2 1 1 1 ... 2 2 3 4 n 2 n 1 n 2 n 1
- ^
- = (^) − (^) − + +^ +^ +^ +
n 1
2n 2 n 2 n 2 n 1 2 n 2 n 1 n 1 n 2
n 1 2 .n 1
n 1 n 2
Binomial Theorem 20.
Q.
n 2 n r r 0
r. C
=
∑
Sol.
(M 1)
n n n r.n n 1 n 1 n 1 r r 1 r 1 r 1 r 0 r 1 r 1
r.( C ) n r. C n (r 1) C C − − − − − − = = =
∑ =^ ∑ =^ ∑ −^ +
n n n 2 n 1 r 2 r 1 r 2 r 1
n(n 1) C n C − − − − = =
− (^) ∑ + ∑
= n(n–1) 2 n–
= n. n– [(n – 1) + 2] = n. n– (n + 1) = n(n + 1)2n–
(M 2)
(1 + x) n = C 0 + C 1 x + C 2 x 2
Differentiating
n(1 + x) n– = C 1 + 2C 2 x + 3C 3 x 2 +…
nx(1 + x) n– = C 1 x + 2C 2 x 2
Differentiating
n[(1 + x) n–
- x.(n – 1) (x + 1) n– ] = C 1 + 2 2 C 2. x + 3 2 C 3 x 2 +…
put x = 1
n[ n–
- (n – 1) 2 n– ] = 1 2 C 1 + 2 2 .C 2 + 3 2
. C 3 + …
n 2 n r r 0
r. C
∑ = n.
n– (2 + n – 1) = n(n + 1) n–
Q.
Show that: S = C 0
+ 3.C
1
+ 5.C
2 +…+ (2n + 1)C n = (n+1).2n
Sol.
S = 1. C 0 + 3.C 1 + 5.C 2 +…+ (2n + 1) Cn
- S = (2n + 1) Cn + (2n – 1) Cn–1 +…+ 1.C 0
2S = (2n + 2) [C 0
+ C
1
+…+ C
n
]
S =
2(n 1) (^) n 2 2
×
∴ S = (n + 1)2n.
