BINOMIAL THEOREM
Bionmial Theorem (For Positive Integer) :
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If x and a are two real positive quantities and n is a positive integer, then the expansion is given by
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(x+a)n = nC
xnao + nC
xn-1a + nC
xn-2a2 +…nC
xn-r ar +… +nC
xoa
- In this expression nC , nC , nC , nC are called binomial
o l 2 n
- In the above expansion, there are (a+l)
- In this expression, the sum of exponents of x and a is always
- The coefficients of equidistant terms from begining and the end are equal, e.
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n n
r n-r
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- In the above expansion the general term Tr+l
- The middle term in the above expansion is
- If n is even the middle term is
= nC
xn-l ar
n
— + 1
2
th
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i.e. only one = nC
xn/2 an/2
- If n is odd then there are two middle terms which are
n + 1
——-
th n + 3 th
and ——–
terms.
2 2
n + 1
——-
th
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term = nC xn+1 an-1
2 2
2 2
n + 3 th
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and———– term = nC
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xn+1 an+1
2
2 2
Binomial Theorem for any Index : This theorem states that (1 + x)n nx n(n – l) n(n – l)(n – 2)
= 1 + +
x2 +
x3 +…
1! 2! 3!
Where n can be positive or negative. The greatest term in the expansion of (l + x)n is
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Tr + 1 nC xr n!(r – 1)!(!(n – r + 1)! n – r + 1
= = x = x.
Tr nC xr-1 r!(n -r)! n! r
r-1
Tr + 1
We can find out——— ³ 1 or Tr+1 ³ Tr where r is a positive integer
Tr
The greatest term in the expansion of (x + a)n is Tr + 1 (n – r + 1)a
=
Tr r x
Term independent of x in the expansion of (x + a)n. Let Tr+l be independent of x.
Equate to zero and find the value of r.
Modifications of Binomial Expansion :
1
— {(l + a)n + (l – a)n} = nC
+ nC a2 + nC
a4 +…
2
1
— {(l + a)n – (l – a)n} = nC
o
+ nC
2 4
a3 + nC
a5 +…
la 3 5
2
Important Properties of Binomial Coefficients :
Bionomial coeff. are written as Co
+ C2
+ C4
+ … = C1
+ C3
+ C5
+… = 2n-1 and
Co + C1 + C2 + C3 + … + Cn = 2
n
where nC
n! n(n – 1)(n – 2)…(n – r + 1)
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= =
r!(n – r)! r!
- C1 + 2C2 + 3C3 + … + nCn
= n. 2n-1
- C1 – 2C2 + 3C3-… = 0
- C0 + 2C1 + 3C2 + … + (n+1) Cn
= (n+2)nn-1
- C0 Cr + C1Cr + 1 + … Cn-rCn (2n)!
=
(n – r)!(n + r)
(2n)
- C 2 + C 2 + C 2 + … + C 2 = ——
0 1 2
n
(n!)2
- C 2-C 2 + C 2-C 2 +…= 0, if n is odd
0 1 2 3
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= (-1)1/2.nC if n is even.
The general term in the expansion of (1 + x)n is given by n(n – 1)…(n – r + 1)
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Tr + 1 = x
r!
Expansions for n = -1, -2 are : (1+x)-1 = 1-x + x2-x3 +… +(-1)r xr + … to ¥
(1-x)-1 = 1 + x + x2 + x3 +…+ xr + … to ¥
(1+x)-2 = 1-2x+3×2-…+(-1)r (r + 1) xr +…to ¥
(1-x)-2 = 1 + 2x + 3x2 + … + (r + 1) xr + …to ¥
(1+x)-3 = 1 – 3x + 6x2 – 10x3 + …
(r + 1)…(r +2)
+(-1)r ——————- xr +… (1-x)-3 = 1 + 3x + 6x2 + 10x3 + …
2!
(r + 1)…(r + 2)
+ xr + …
2!
Some Results :
- If coefficient of rth, (r + 1)th and (r + 2)th terms in the expansion of (1 + x)n are in H.p., then n + (n – 2r)2 = 0
- If coefficient of rth (r + 1)th, and (r + 2)th terms in (l + x)n are in P, then n2-n (4r + 1) + 4r2-2 = 0
Bionomial theorem is of great importance in algebra.
