Binomial Theorem for a Positive Integral Index
Binomial Theorem is one of the easiest and important chapters of Algebra in the syllabus of IIT JEE. The beginners sometimes find it difficult as this topic is very new to them Binomial. The other reason is that the concepts of Combinations are also used in Theorem. This chapter can be said to be one of the easiest as it has very few twists and turns.
The chapter begins with the Introduction to Binomial Theorem which is followed by the Properties of the Binomial Theorem. Binomial Coefficients are the most important topic of Binomial theorem. The sum of the Binomial coefficients, the coefficient of a particular term, greatest Binomial coefficient etc. has been discussed at length. However, some important results have been given at the end which is supplemented by the application of Binomial Expression. Solved examples as usual are very useful as they can reappear in the examination with slight modification.
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What is the Binomial Theorem for a positive integral?
The binomial theorem explains the way of expressing and evaluating the powers of a binomial. This theorem explains that a term of the form (a+b)^{n} can be expanded and expressed in the form of ra^{s}b^{t}, where the exponents s and t are nonnegative integers satisfying the condition s + t = n. The coefficient r is a positive integer. The terms involved are called binomial coefficients and since it is for positive indices it expands only the positive powers.
The general binomial expansion for any index is given by
(x+y)^{n} = ^{n}C_{0}x^{n}y^{0} + ^{n}C_{1}x^{(n1)}y^{1} + ^{n}C_{2}x^{(n2)}y^{2 }+ …….. + ^{n}C_{(n1)} x^{1}y^{(n1)} + ^{n}C_{n}x^{0}y^{n}.

The topic is quite wide and includes various heads like:
We shall be discussing these topics here in brief as they have been discussed in detail in the coming sections.
Some of the properties of Binomial Expansion of the term (x+y)^{n}:
1. Any expansion of this form has (n+1) terms.
2. The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is
C (n, r) = C (n, nr) which gives C (n, n) C (n, 1) = C (n, n1) C (n, 2) = C (n, n2).
3. The indices of x and y always sum to n.
4. Such an expansion always follows a simple rule which is
 The subscript of C i.e. the lower suffix of C is always equal to the index of y and
 Index of x = n – (lower suffix of C).
5. In the expansion of the term (x+y)^{n}, the (r+1)th term is called the general term and is given by
T_{(r+1)} = ^{n}C_{r}x^{nr}y^{r}

Middle term: The middle term of the binomial coefficient depends on the value of n. There can be two different cases according to whether n is even or n is odd.
 If n is even, then the total number of terms are odd and in that case there is a single middle term which is (n/2 +1)th and is given by
^{n}C_{n/2 }a^{n/2 }x^{n/2}

 On the other hand, if n is odd, the total number of terms is even and then there are two middle terms [(n+1)/2]th and [(n+3)/2]th which are equal to
^{n}C_{(n1)/2} a^{(n+1)/2}x^{(n1)/2}

^{ }
^{n}C_{(n+1)/2} a^{(n1)/2}x^{(n+1)/2}

 The binomial coefficient of the middle term is the greatest binomial coefficient of the expansion.
Illustration: In the binomial expansion of (ab)^{n}, n ≥5, the sum of the 5^{th} and 6^{th} terms is zero. Then find the value of a/b.
Solution: The sum of the 5^{th} term is given by
T_{5} = ^{n}C_{4}a^{n4}(b)^{4}
The sum of the 6^{th} term is given by
T_{6} = ^{n}C_{5}a^{n5}(b)^{5}
It is given in the question that T_{5} + T_{6} = 0.
This gives a/b = (n4)/5.
Illustration: Prove that
C_{0} – 2^{2}C_{1} + 3^{2}C_{2}  ….. + (1)^{n}(n+1)^{2}C_{n} = 0, n > 2, where C_{r }= ^{n}C_{r}.
Solution: We know that by the binomial theorem
(1+x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + …. + C_{n}x^{n}
Multiplying it by x we get,
x(1+x)^{n} = C_{0}x + C_{1}x^{2} + C_{2}x^{3} + ……. + C_{n}x^{n+1}
Differentiating both sides we get,
(1+x)^{n} + nx(1+x)^{n1} = C_{0} +2C_{1}x +3C_{2}x^{2} + ….. + (n+1)C_{n}x
Again multiplying by x we get,
x(1+x)^{n} + nx^{2}(1+x)^{n1} = C_{0}x + 2C_{1}x^{2} + 3C_{2}x^{3} + ……. + (n+1)C_{n}x^{n+1}
Hence, (1+x)^{n} + nx(1+x)^{n1} + 2 nx(1+x)^{n1} + n(n1)x^{2}(1+x)^{n2} = C_{0} +2^{2}C_{1}x +3^{2}C_{2}x^{2} + ….. + (n+1)^{2}C_{n}x^{n}
Putting x = 1, we get
0 = C_{0} – 2^{2}C_{1} + 3^{2}C_{2}  ….. + (1)^{n}(n+1)^{2}C_{n}
Binomial Theorem is important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic. The chapters of Binomial Theorem, Permutations and Combinations, and Probability together fetch 1020 marks varying from one examination to the other. askIITians offers a unique platform where the students are free to ask their questions on topics like sum of binomial coefficients, applications of binomial theorem and greatest coefficient binomial theorem.
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