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Mathematics
Session
Binomial Theorem Session 2
Session Objectives
Session Objective
1. Properties of Binomial Coefficients
2. Binomial theorem for rational index
— General term
— Special cases
3. Application of binomial theorem
— Divisibility
— Computation and approximation
Properties of Binomial Coefficients
1. The sum of the binomial coefficients in the expansion of (1+x)n is 2n i.e.
n0 1 nC C ... C 2
where Cr = nCr
Proof:
n n n n 2 n n0 1 2 n1 x C C x C x ... C x
Put x = 1 both sides we get
n n n n n0 1 2 nC C C ...... C 2
For example 4C0 + 4C1 + 4C2 + 4C3 + 4C4
= 1 + 4 + 6 + 4 + 1 = 16 = 24
Properties of Binomial Coefficients
2. Sum of coefficients of the odd terms = Sum of the coefficients of the even terms in (1+x)n = 2n-1 i.e.
n–10 2 4 1 3 4C C C ... C C C ... 2
Proof: n n n n n0 1 n1 x C C x ... C x
Put x = –1 in above, we get
nn n n n0 1 2 n0 C C C ... C 1
n n n n n n0 2 4 1 3 5C C C ... C C C ...
As sum of all the coefficients is 2n
n n n0 22 C C ..... 2
n–10 2 4 1 3 4C C C ... C C C ... 2
n
rnr
r 0
C 1 0
Properties of Binomial Coefficients
3. n n 1 n 2r r 1 r 2
n n n 1C C . . C and so on
r r r 1
Proof:
n n–1
r r–2
n. n – 1 !n! nC C
r! n – r ! r. r – 1 ! n – r ! r
n–2r–2
n n – 1 n – 2 ! n n – 1. . C
r r – 1 r – 2 ! n – r ! r r – 1
For example: 4 3 22 1 0
4 4 3C . C . . C 6
2 2 1
Class Exercise - 1If C0, C1, C2,... denote the binomial coefficients in the expansion of (1+x)n then prove that
2 2 2 2 n–21 2 3 n1 C 2 C 3 C ... n C n n 1 2 .
Solution :
2 2 2 2 n–21 2 3 n1 C 2 C 3 C ... n C n n 1 2
n n2 n 2 n–1
r r–1r 1 r 1
nLHS r C r . . C
r
n
n–1r–1
r 1
n r C
n
n–1r–1
r 1
n r – 1 1 C
n n2 n n 1
r r 1r 1 r 1
r C n r 1 1 C
n n
n–1 n–1r–1 r–1
r 1 r 1
n r – 1 C n C
n n
n–2 n–1r–2 r–1
r 2 r 1
n – 1n r – 1 C n C
r – 1
n n
n–2 n–1r–2 r–1
r 2 r 1
n n – 1 C n C
Solution Cont.
n n
n–2 n–1r–2 r–1
r 2 r 1
n n – 1 C n C
n–2 n–1 n–2n n – 1 2 n.2 n n 1 2
Class Exercise - 2
Solution :
2 2 20 1 2 3
n 2n
C 2 C 3 C 4 C
... 1 n 1 C 0, n 0.
n n
r 2 r 2r r
r 0 r 0
LHS –1 r 1 C –1 r 2r 1 C
n
rr
r 0
–1 r r – 1 3r 1 C
n n nr r r
r r rr 0 r 0 r 0
1 r r 1 C 3 1 r C 1 C
Solution Cont.
n n nr r r
r r rr 0 r 0 r 0
1 r r 1 C 3 1 r C 1 C
n nr rn 1
r 1 rr 1 r 0
n3 1 r. . C 1 C
r
n r n 2
r 2r 2
n n 11 r r 1 . C
r r 1
n n
r rn–2 n–1r–2 r–1
r 2 r 1
n n – 1 –1 C 3n –1 C 0
= 0 + 0 + 0 = 0
Binomial theorem for Rational Index
Let n be a rational number and x be real number such that |x| < 1, then
n 2 3n n 1 n n 1 n 21 x 1 nx x x ...
2! 3!
rn n 1 n 2 ... n r 1... x ...
r!
Conditions of validity: if n is not a whole number
i) |x| < 1 ii) Number of terms is infinite
General Term in (1+x)n, n Q
rr 1
n n 1 n 2 ... n r 1T x
r!
Expansion of (a + x)n for rational n
Case1: a
1x
n
n n ax a x 1
x
2 3n n n 1 n n 1 n 2a a a
x 1 n. ...x 2! x 3! x
Case2: x
1a
n
n n xx a a 1
a
2 3n n n – 1 n n – 1 n – 2x x x
a 1 n. ...a 2! a 3! a
Class Exercise - 6
Solution :
Write the first four terms in the expansion of
For what values of x is this expansion is valid? Also, find the general term in this expansion.
1
2 24 5x
–1
–1–1 2 22 22 5x4 – 5x 4 1–
4
22 2–1 –1
– 11 –1 –5x –5x2 2
(12 2 4 2! 4
32–1 –1 –1
– 1 – 2–5x2 2 2
...)3! 4
2 64
3
1 5x 75 1.3.5 125x1 x . ...
2 8 128 642 .3!
62 41 5 75 625 x
x x ...2 16 256 2048
Solution Cont.
Validity if2
25x 41 x
4 5
2 4 2x x
5 5
–2 2x
5 5
General termr2
r 1
–1 –1 –1– 1 ... – r 1
1 –5x2 2 2T
2 r! 4
r2
r
–1 –3 –5 ... –1– 2r 21 –5x.
2 42 r!
rr 2
r
–1 1.3.5... 2r – 11 –5x.
2 42 r!
r2r
r 1
1.3.5... 2r – 1 5x
4r! 2
Special Cases
n 2
3
n n 1i) 1 x 1 nx x
2!
n n 1 n 2x ...
3!
r rr 1
n n 1 ... n r – 1T –1 x
r!
n 2 3n n 1 n n 1 n 2ii) 1 x 1 nx x x ...
2! 3!
rr 1
n n 1 ... n r – 1T x
r!
Special Cases
n 2
3
n n 1iii) 1 x 1 nx x
2!
n n 1 n 2x ...
3!
r rr 1
n n – 1 ... n – r 1T –1 x
r!
Application
n n n 2 n n–1 n n1 2 n–1 n1 x 1 C x C x ... C x C x
n n n n n–11 2 n1 x – 1 x C C x ... C x
= Multiple of x = M(x)
n 21 x 1 nx M x
n 2 3n n – 11 x – 1– nx – x M x
2
Conclusion: The number by which division is to be made can be x or x2 or x3, but the number in the base is always expressed in the form of 1 + kx.
I Division
Class Exercise - 9
Solution :
Which of the following expression is divisible by 1225?
(a) (b)
(c) (d)
2n6 – 35n – 1 2n6 – 35n 12n6 – 35n 2n6 35n – 1
21225 35
n2n n6 36 1 35
n 2 n n2 n1 35n C 35 ... C 35
2n 2 n n n n–22 3 n6 – 35n – 1 35 C C 35 ... C 35
= 1225 k2n6 – 35n – 1 is divisible by 1225.
Application
I Divisionn nx – y is divisible by x – y for all
positive integer n
Proof:
nn n nx – y x – y y – y
n n–1n n n n1 nx – y C x – y y ... C y – y
n n–1n n n–11 n–1x – y C x – y y ... C x – y y
As each term is divisible by x – y,
xn - yn is divisible by x – y
Application
II Computation and Approximation
Find 99993 exactly
Solution :
333 49999 10000 1 10 1
3 24 3 4 3 41 210 C 10 C 10 1
12 8 410 3.10 3.10 1
= 1000000000000 - 300000000 +30000 -1
= 999700029999
Class Test
Class Exercise - 3
Solution :
31 2 n
0 1 2 n–1
n n 1CC C C2 3 ... n .
C C C C 2
n n
r r–1n! n – r 1n! n – r 1
C Cr! n – r ! r r – 1 ! n – r 1 ! r
nr r
nr–1r–1
C Cn – r 1
r CC
LHS = n n n
r
r–1r 1 r 1 r 1
C n – r 1r r. n – r 1
C r
n n 1 n n 1n 1 n –
2 2
= RHS
Class Exercise - 4
Solution :
2 2 2 n0 1 n 2
2n ! 1.3.5... 2n – 1C C ... C 2
n!n!
2 2 20 1 nC C ... C
0 0 1 1 n nC .C C .C ... C .C
n no n 1 n–1 n 0 r n–rC .C C .C ... C .C as C C
n n0 1 n1 x C C x ... C x
n n–1n n–1 0C x C x ... C
n n n n0 1 n n 01 x . 1 x C C x ... C x C x ... C
Class Exercise - 4
Solution :
2 2 2 n0 1 n 2
2n ! 1.3.5... 2n – 1C C ... C 2
n!n!
n n n n0 1 n n 01 x . 1 x C C x ... C x C x ... C
Compare the coefficient xn of both sides
2n 2 2 2n 0 1 nC C C ... C
2 2 2 2n0 1 n n
2n ! 1.2.3...2nC C ... C C
n! n! n! n!
n1.3.5.7...(2n – 1).2 .1.2...nn! n!
n1.3.5.7...(2n – 1).2n!
Class Exercise - 5
Solution :
n 11 2 n
0C C C 2 – 1
C ...2 3 n 1 n 1
LHS = n n
r
r 0 r 0
C n!r 1 r! n – r ! r 1
n
r 0
n 1 !1n 1 r 1 ! n – r !
n
n 1 n 1r 1
r 0
1 1C 2 – 1
n 1 n 1
= RHS
Class Exercise - 7
Solution :
Find the coefficient of x4 in the expansion
of Also find the coefficient of xr
and find its expansion.
21– x
.1 x
2
2 –21– x1– x 1 x
1 x –221– 2x x 1 x
2 2 32 2 1 2 3 41 x (1 2x x x
2! 3!
42 3 4 5x ...)
4!
2 3 41– 2x 3x – 4x 5x – ... r r
r 0
–1 r 1 x
Solution Cont.
2
2 2 3 41– x1– 2x x 1– 2x 3x – 4x 5x ...
1 x
Coefficient of x4 is 1.5 – 2.(–4) + 1.3 = 5 + 8 + 3 = 16
Coefficient of xr is r r –1 r–2–1 r 1 – 2 –1 r –1 r – 1
r r–1 r 1 2r r – 1 –1 .4r
2
r r
r 1
1 x1 4 1 rx
1 x
Class Exercise - 8
Solution :
When x is so small that its square and higher powers may be neglected, find
the value of
–5
3
21 x 4 2x
3
4 x
1–5 –5
2
33 322
2 2 x1 x 4 2x 1 x 2 1
3 3 2
4 x x4 1
4
2 1 x1– 5 x 2 1 .
3 2 23 x
8 1 .2 4
As terms involving x2, x3, neglected
Solution Cont.
2 1 x1– 5 x 2 1 .
3 2 23 x
8 1 .2 4
10 x1– x 2
3 28 3x
–117x3 – 8 3x
6
11 17x 3x
3 18 6 8
1 17x 3x3 1
8 6 8
1 9x 17x3 – –
8 8 6
1 27 683 x
8 24
1 95x3 –
8 24
Thank you