Binomial Theorem Class 11 Notes Mathematics Chapter 8 - CBSE

Chapter : 8

What Are Binomial Theorem ?

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    An algebraic expression is built from the constants, variables and finite number of algebraic operations i.e., addition, subtraction etc. An algebraic expression that contains only two terms is called binomial.

    The binomial theorem is an algebraic formula by which we can find the value of any positive integer exponent of a binomial.

    In other words, binomial theorem, helps to expand (x + y)n for any positive integral index n. The binomial theorem gives us a simple way to simplify such large terms.

    History Of Binomial Theorem

    Since the 4th century BC humankind has been able to demonstrate the binomial theorem. This cube binomial was also utilised in the 5th century AD. Halyudth, an Indian mathematician established this approach in the 10th century AD utilishing Pascal's triangle.

    In the 12th century, this theorem was explicity established Mathematicians advanced these findings until Sir Issac Newton, in 1665, expanded the binomial theorem to all exponents.

    Pascal Triangle

    Pascal's triangle, in algebra, a triangular arrangements to numbers that gives the coefficients in the expansion of any binomial expression such as (x + y)n . It is named for the 17th century French Mathematician Blaise Pascal.


    Binomial Theorem For Positive Integral Index

    The binomial theorem helps to expand any positive integral power of the binomial expression.

    If a and b are real numbers, then for all n ∈ N.

    (a + b)n = nC0 an + nC1 an–1b + nC2 an–2b2 +...+ nCn–1 abn–1 + nCnbn

    Proof of Binomial Theorem

    Proof: This theorem is proved by using the principle of mathematical induction.

    Let P(n) be the statement

    (a + b)n= nC0 an + n C1 an–1 b + nC2an–2b2 + ... + nCn–1 abn–1 + nCn bn

    Then, the statement P(1) is :

    (a + b)1 = 1C0 a1 + 1C1 a0 b = (a + b), which is true.

    Let P(m) be true. Then,

    (a + b)m = mC0 am + mC1 am–1 b + mC2 am–2 b2 + ... + mCm–1 abm–1 + mCmbm ...(i)

    Multiply both sides of (i) by (a + b), we get

    (a + b)m+1 = mC0 am+1 + mC0amb + mC1 amb + mC1 am–1b2 + mC2 am–1b2 + mC2 am–2b3 +...+ mCm−1 a2 bm−1 + mCm–1abm+ mCmabm + mCmbm + 1

    = mC0 am+1 + (mC0 + mC1)amb + (mC1 + mC2)am–1b2 +... + (mCm–1 + mCm)abm + mCmbm+1

    = m+1C0 am+1 + m+1C1 amb + m + 1C2 am−1b2 + ... + m+1Cabm + m+1Cm+1 bm+1

    $$\begin{bmatrix}\therefore\space ^{m}\text{C}_{0} = 1 = ^{m+1}\text{C}_{0},\\^{m}\text{C}_{m}1= 1 =^{m+1}\text{C}_{m+1}\\\text{and}\space ^{m}\text{C}_{r-1} + ^{m}\text{C}_{r} = \space^{m+1}\text{C}_{r}\end{bmatrix}$$

    This shows that P(m + 1) is true, whenever P(m) is true. Hence, by the principle of mathematical induction the theorem is true for all n ∈ N. So, the above result can be written as,

    $$(a+b)^{n} =\sum^{n}_{r =0}\text{C}_{r}a^{n-r}b^{r}$$

    Binomial Theorem By Positive Integral Indices

    Observe the following identities :

    • (a + b)0 = 1
    • (a + b)1 = a + b
    • (a + b)2 = a2 + 2ab + b2
    • (a + b)3 = a3 + 3a2 b + 3ab2 + b3
    • (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

    From this we conclude that

    • The exponent of a in the expansion of (a + b)n is decreasing by 1 from n to 0.
    • The exponent of b in the expansion of (a + b)n increasing by 1 from 0 to n.
    • The sum of the exponents of a and b in the expansion of (a + b)n is always n.
    • The number of terms in the expansion of (a + b)n is always n + 1.
    • The coefficients of terms in above expression are actually making a Pascal's triangle.

    Similarly, (a + b)5 , (a + b)6 can be expanded using the Pascal's triangle. But working with Pascal's triangle can be very time consuming if the exponent is a very large number (say 100).

    That is why, binomial theorem is used for the expansion of (x + y)n for any positive integral index.