# Binomial  Theorem

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. If   x, y ∈ R and n∈N,  then

(x + y)n = nC0 xn + nC1 xn-1 y + nC2 xn-2y2 + ….. + nCrxn-r yr + ….. + nCnyn =nCr xn – r yr

This theorem can be proved by Induction method.

OBSERVATIONS :

(i)  The number of terms in the expansion is (n + 1) i.e. one or more than the index.

(ii) The sum of the indices of x & y  in each term is n.

(iii) The binomial coefficients of the terms nC0, nC1……..  equidistant from the beginning and the end are equal.

Important Terms In The Binomial Expansion Are :

(i)  General term

(ii) Middle  term

(iii)  Term  Independent  of  x

(iv) Numerically  greatest  term

(i)  The  general  term  or  the (r + 1)th  term  in  the  expansion  of  (x + y)n  is  given  by ;

Tr+1 = nCr xn-r .yr

(ii) The  middle  term(s)  is  the  expansion  of  (x + y)n  is (are) :

(a) If  n  is  even, there  is  only  one  middle  term  which  is  given  by  ;

T(n+2)/2 = nCn/2 . xn/2. yn/2

(b) If  n  is  odd, there  are  two  middle  terms  which  are :

T(n+1)/2    &    T[(n+1)/2]+1

(iii)      The term independent of x is a term which does not have x;  Hence find the value of r  for which the exponent of x  is zero.

(iv)      To find  the  Numerically greatest  term in  the  expansion  of  (1 + x)n , n∈N find                                                      Tr+1/T= Cr/Cr-1 = (n-r+1)/r .

Put the absolute value of x & find the value of r  Consistent  with  the  inequality                                                                           Tr+1/T> 1

Note that the  Numerically greatest term in the expansion of  (1 – x)n , x > 0, n∈N  is the same as the greatest term in  (1 + x)n .

If  (√A+B)n = I + f,  where   I and  n  are  positive  integers,  n  being  odd  and  0 < f < 1, then
(I + f) . f = Kn  where  A – B2 = K > 0  &  √A-B<0.

If n  is an even integer,  then  (I + f) (1 – f) = Kn.

Binomial Coefficients

(i)  C0 + C1 + C2 + ……. + C= 2n

(ii) C0 + C2 + C4 + …….= C1 + C3 + C5 + …….= 2n-1

(iii) C0² + C1² + C2² + …. + Cn² = 2nCn

(iv) C0.Cr + C1.Cr+1 + C2.Cr+2 + … + Cn-r.Cn = (2n)!/(n+r)(n+r-1)

Important Note:

(2n)! = 2n.n! [1. 3. 5 …… (2n – 1)]

Binomial Theorem  For  Negative  Or  Fractional  Indices :

If n∈Q, then (1+x)n =1+nx+n(n-1)/2!+n(n-1)(n-2)/3!+……….    Provided | x | <  1.

When the  index  n  is  a  positive  integer  the  number  of  terms  in  the  expansion of

(1 + x)n  is  finite  i.e.  (n + 1)  &  the  coefficient  of  successive  terms  are  :

nC0 , nC1 , nC2 , nC3 ….. nCn

When the index is other than a  positive integer such as negative integer or fraction,  the number of terms in the expansion of  (1 + x)n  is infinite and the symbol  nCr  can not be used to denote the  Coefficient of the general term.

The following expansion  should  be  remembered  (modulas x < 1).

(a)  (1 + x)-1 = 1 – x + x2 – x3 + x4 – ….

(b)  (1 – x)-1 = 1 + x + x2 + x3 + x4 + ….

(c)  (1 + x)-2 = 1 – 2x + 3x2 – 4x3 + ….

(d)  (1 – x)-2 = 1 + 2x + 3x2 + 4x3 + …..

The expansions in ascending powers of x  are only valid if x  is ‘small’.If x  is large |x | > 1  then we may find it convenient to expand in powers of  1/x,  which then will be small.

Approximations :

(1 + x)n = 1 + nx +n(n-1) x²/2! +n(n-1)(n-2)  x3/3!………..  If x < 1,  the terms of the above expansion go on decreasing and if x  be very small, a stage may be reached when we may neglect the terms containing higher powers of x  in the expansion.Thus, if  x  be so small that its squares and higher powers may be neglected then

If x < 1,  the terms of the above expansion go on decreasing and if x  be very small, a stage may be reached when we may neglect the terms containing higher powers of x  in the expansion.Thus, if  x  be so small that its squares and higher powers may be neglected then

(1+x)n=1+nx,  approximately.

This is an approximate value of  (1 + x)n.

Exponential Series:

(i)  ex =1+x/1!+x3/3!+x4/4!……

where x may be any real or complex & e=

(ii)  ax=1+(xloga)/1!+ (xloga)2/2!+ (xloga)3/3!+ (xloga)4/4!+……….where a > 0

e is an irrational number lying between 2.7 & 2.8. Its value correct up to 10 places of the decimal is 2.7182818284…..

Logarithms to the base ‘e’ are known as the Napierian system, so named after Napier, their inventor. They are also called  Natural Logarithm.

Logarithmic  Series :

(i)   ln(1-x)=x-x2/2+x3/3-x4/4……….    where -1 < x  1

(ii)  ln(1-x)=-x-x2/2-x3/3-x4/4……….     where  -1 x < 1

Important Notes:

(a) ln 2=1-1/2+1/3-1/4………..

(b) eln x = x

(c)  ln2 = 0.693

(d) ln10 = 2.303

Here I am posting a Pdf of Binomial theorem’s 100 questions. you can download and solve them for your practice.

Binomial Theorem.pdf.pdf

## How to find sum and product of zeros of equations

In my previous post on IB Mathematics, I have discussed how to solve a quadratic polynomial using Quadratic formula. Here I will tell you about different relationships based on sum and product of quadratic polynomial, cubic polynomial and bi-quadratic polynomials.

ax2 + bx + c = 0

Sum of the roots = −b/a

Product of the roots = c/a

If we know the sum and product of the roots/zeros of a quadratic polynomial, then we can find that polynomial using this formula

x2 − (sum of the roots)x + (product of the roots) = 0

## Cubic:

Now let us look at a Cubic (one degree higher than Quadratic):

ax3 + bx2 + cx + d=0

if α, β and γ are the zeros of this cubic polynomial then

α+β+γ=-b/a

αβ+βγ+γα=c/a

αβγ=-d/a
If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:

x³-(α+β+γ)x²+(αβ+βγ+γα)x-(α+β+γ)=0

If we are given a bi-quadratic polynomial with degree 4 like:

ax 4+bx³+cx²+dx+e=0

and its roots/zeros are α, β, γ, and δ then

α+β+γ+δ=-b/a

αβ+αγ+αδ+βγ+βδ+γδ=c/a

αβγ+βγδ+γδα+αβδ=-d/a

αβγδ=e/a
using these formulas of sum and product of zeros of polynomials, we can find a lot of relationships in zeros of polynomials. Usually, we are asked to find these types of relationships in zeros.
Question: If α and β are the zeros of polynomial x²-px+q=0 then find the following relationships.
i) 1/α+1/β   ii) α²β+αβ²   iii) α²+β²    iv) α/β+β/α   v) α³+β³
Ans: To find these relationships we convert every value either in sum (α+β)  or in the product (αβ) of zeros. For this conversion, we use following Mathematical Tricks
1)Try to take common
2) Try to take L.C.M
3) Try to make a Perfect Square
4) Use algebraic identities wherever required
If we use above steps properly, in most of the cases we are able to convert everything either in sum or in product of zeros
If we compare the given equation with std. form ax²+bx+c=0 then
a=1, b=-p and c=q
sum of zeros α+β=-b/a=-(-p/a)=p

product of zeros    αβ=c/a=q/1=q                                                     (i) 1/α+1/β=β+α/αβ              [By L.C.M]

=p/q
(ii) α²β+αβ²  = αβ(α+β)               [By common]
=q.p
(iii) α²+β² = (α²+β²+2αβ)-2αβ          [add and subtract 2αβ, make it a perfect square ]

=(α+β)²-2αβ

=p²-2q
(iv) α/β+β/α = β²+α²/βα          [By taking L.C.M]

we have already found β²+α² that is p²-2q so
α/β+β/α=(p²-2q)/q

(v) (α+β)³=(α+β)³-3αβ(α+β)                [Direct algebraic identity]

=p³-3qp