Binomial Theorem

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.

<img src="binomial theorem.png" alt="binomial theorem">


(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=  {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{n}} \right)^n}

(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.
You can express your views and ask your doubts in the comment section

 Binomial Theorem.pdf.pdf 




Calculus (Part-2)-geometric explanation of differentiation

Online maths tutors like the following  concept very much

Geometric explanation of differentiation-  If we find the derivative of function f(x) at x = x0 then it is equal to the slope of the tangent  to the graph of given function f(x) at the given point [(x0, f(x0))].

But what is a tangent line?

It is not merely a simple line that joins the graph of given function at one point.

It is actually the limit of the secant lines joining points P = [(x0, f(x0)] and Q on the graph of f(x) as Q moves very much close to P.

The tangent line contacts the graph of given function at given point [(x0, f(x0)] the slope of the tangent line matches the direction of the graph at that point. The tangent line is the straight line that best approximates the graph at that point.

<img src="derivative and slope.png" alt="derivative and slope">


As we are given the graph of given function, we can draw the tangent to this graph easily. Still, we’ll like to make calculations involving the tangent line and so will require a calculative method to explore the tangent line.


We can easily calculate the equation of the tangent line by using the slope-point form of the line. We slope of a line is m and its passing through a point (x0,y0) then its equation will be

                                                              y − y0 = m(x − x0)


So now we have the formula for the equation of the tangent line. It’s clear that to get an actual equation for the tangent line, we should know the exact coordinates of point P. If we have the value of x0 with us we calculate y0 as

                                                                      y = f(x0)

The second thing we must know is the slope of line

                                                                      m = f’(x0)

Which we call the derivative of given function f(x).



The derivative f’(x0) of given function f at x0 is equal to the slope of the tangent line to

                                        y = f(x) at the point P = (x0, f(x0).

Differentiation Using Formulas- We can use derivatives of different types of functions to solve our problems :


{\rm{(i) D (}}{{\rm{x}}^{\rm{n}}}{\rm{) = n}}{\rm{.}}{{\rm{x}}^{{\rm{n - 1}}}}{\rm{ here n, x}} \in R        \;\left( {ii} \right){\rm{ }}\;D{\rm{ }}\left( {{e^x}} \right){\rm{ }} = {\rm{ }}{e^x}

{\rm{(iii) D(}}{{\rm{a}}^{\rm{x}}}{\rm{) = }}{{\rm{a}}^{\rm{x}}}{\rm{lo}}{{\rm{g}}_e}{\rm{a a > 0}}                {\rm{(iv) D (ln x) = }}\frac{1}{x}

{\rm{(v) D(lo}}{{\rm{g}}_{\rm{a}}}{\rm{x) = lo}}{{\rm{g}}_{\rm{a}}}{\rm{e }}                      {({\rm{vi}}){\rm{D}}({\rm{sinx}}){\rm{ = cosx}}}

{({\rm{vii}}){\rm{D}}({\rm{cosx}}){\rm{ = - sinx}}}                     {({\rm{viii}}){\rm{D(tanx) = se}}{{\rm{c}}^2}x}

(ix)  D (secx) = secx . tanx              (x)  D (cosecx) = – cosecx . cotx   

{\rm{(xi) D (cotx) = - cose}}{{\rm{c}}^2}x              (xii) D (constant) = 0 where D = {\frac{d}{{dx}}}

These formulas are result of differetiation by first principle

Inverse Functions And Their Derivatives :


{\rm{(i) }}\frac{d}{{dx}}({{{\rm Sin}\nolimits} ^{ - 1}}x) = \frac{1}{{\sqrt {{1^2} - {x^2}} }}                 {\rm{(ii) }}\frac{d}{{dx}}({\cos ^{ - 1}}x) = \frac{{ - 1}}{{\sqrt {{1^2} - {x^2}} }}

{\rm{(iii) }}\frac{d}{{dx}}({{{\rm Sec}\nolimits} ^{ - 1}}x) = \frac{1}{{x\sqrt {{x^2} - 1} }}              {\rm{(iv) }}\frac{d}{{dx}}({{{\rm cosec}\nolimits} ^{ - 1}}x) = \frac{{ - 1}}{{x\sqrt {{x^2} - 1} }}

{\rm{(v) }}\frac{d}{{dx}}(Co{t^{ - 1}}x) = \frac{{ - 1}}{{1 + {x^2}}}                  {\rm{(vi) }}\frac{d}{{dx}}({\tan ^{ - 1}}x) = \frac{1}{{1 + {x^2}}}

Theorems On Derivatives:  If u and v are derivable function of x, then,


(i)   \begin{array}{l} {\rm{ }}\frac{d}{{dx}}(u \pm v) = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}\\ \\ \end{array}                    (ii)   \frac{d}{{dx}}k[f(x)] = k\frac{d}{{dx}}f(x)    where K is any constant


(iii)  {\rm{ }}\frac{d}{{dx}}(u.v) = v\frac{{du}}{{dx}} + u\frac{{dv}}{{dx}}    known as  “ Product  Rule ”      (iv)   {\rm{ }}\frac{d}{{dx}}(\frac{u}{v}) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}

known as  “Quotient  Rule ”


(v) If  y = f(u)  &  u = g(x)  then  \frac{{dy}}{{dx}} = \frac{{dy}}{{du}}.\frac{{du}}{{dx}}    “Chain  Rule ”


Logarithmic  Differentiation:  To find the derivative of  : (i) a function which is the product or quotient of a number of functions OR
(ii) a function of the form  {\left[ {f\left( x \right)} \right]^{g\left( x \right)}} where f & g  are both differentiable, it will be found convenient to take the logarithm of the function first & then differentiate. This is called Logarithmic  Differentiation.


Implicit  Differentiation:  (i) In order to find dy/dx,  in the case of implicit functions, we differentiate each term w.r.t.  x  regarding y as a function of x & then collect terms in dy/dx together on one side to finally find dy/dx.

(ii) In answers of dy/dx in the case of implicit functions, both x & y are present.

Parametric  Differentiation: If  y = f(q)  &  x = g(q)  where q  is  a parameter, then

                                                                         \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }}

Derivative Of A Function w.r.t. Another Function-: Let  y = f(x) & z = g(x)

\frac{{dy}}{{dx}} = \frac{{f'(x)}}{{g'(x)}}

Derivatives Of Order Two & Three: Let a function y = f(x) be defined on  an open interval (a, b).  It’s derivative, if it exists on(a, b)  is  a certain function f'(x) [or (dy/dx) or y’ ] &  is  called  the  first derivative of y  w.r.t. x. If it happens that the first derivative has a derivative on (a, b) then this derivative is called the second derivative of y  w. r. t.  x  &  is denoted by f”(x) or (d2y/dx2) or  y”. Similarly, the 3rd order derivative of y w. r. t.  X, if it exists, is defined by    It is also denoted by f”'(x) or y”’.

All online maths tutors suggest solving fair amount of questions based on these concepts


Example-Find the tangent line to the following function at z=3 for the given function


We can find the derivative of the given function using basic differentiation as discussed in the previous post            

We are already given that z=3 so

Equation of tangent line is

                                                    y − y0 = m(x − x0)      here y0=R(3)=√7

Putting these values we get equation of tangent line


In my online maths tutors series, I will discuss Application of Derivatives


Example-  Differentiate the following function    


Ans: We can apply quotient rule in this questions


You can try these questions in the meantime

 28 – Differentiation (09-10-15) with answers.pdf



<img src="demo.png" alt="demo">



IB Mathematics Tutors-Part 1 (Calculus)

In my IB Mathematics Tutors series, I will explain different topics taught at HL and SL levels of IB Mathematics. Calculus is the first one.

<img src="ib mathematics tutors.jpg" alt="ib mathematics tutors">

If we want to understand the importance of Calculus in IB Diploma Programme, We should have a look at the number of teaching hours recommended for it. It’s 40 hours in SL(out of 150 total hours) and 48 (out of 240 total hours) hours in HL. This makes calculus the most important topic for IB Mathematics Tutors as well as for IB students.

What is Calculus- Calculus is an ancient Latin word. It means ‘small stones’ used for counting. In every branch of Mathematics, we study of something specific, like in Geometry, we study about shapes, in Algebra, we study about arithmetic operations, in coordinate, we study about locating a point. In calculus, we do the mathematical study of continuous change. It mainly has two branches- Read more

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.

<img src="ib mathematics.png" alt="ib mathematics">


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


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

If we know these relationships of polynomials then we cal also calculate the polynomial using this formula:


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
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]

(ii) α²β+αβ²  = αβ(α+β)               [By common]
 (iii) α²+β² = (α²+β²+2αβ)-2αβ          [add and subtract 2αβ, make it a perfect square ]
(iv) α/β+β/α = β²+α²/βα          [By taking L.C.M]
we have already found β²+α² that is p²-2q so

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

You can further read about quadratics in PDF(quadratics ) given here. There are a lot of practise questions given in this PDF
<img src="demo.png" alt="demo">