## Binomial Coefficients and their Properties,Application of Binomial Theorem for Approximation

Subject: Mathematics

#### Overview

This note helps you learn about binomial coefficients and their properties as well as the applications of the Binomial theorem in approximation.

#### Binomial Coefficients and their Properties

For the expansion of binomial coefficients, we consider the expansion of (1+x)n, where n is a natural number. Then, From the Binomial Theorem,

$$\;(1+x)^n\;=\;^nC_0\;+\;^nC_1x\;+\;^nC_2x^2\;+\;.\;.\;.\;.\;.\;+\;^nC_nx^n$$

$$\;i.e.\;(1+x)^n\;=\;1\;+\;nx\;+\;n(n-1)\frac{x^2}{2!}\;+\;\frac{n(n-1)(n-2)x^3}{3!}\;+\;.\;.\;.\;.\;.\;.\;(1)$$

NOTE:$$In\;the\;places\;of\;^nC_0\;,\;^nC_1\;,\;^nC_2\,;.\;.\;.\;\;we\;may\;write\;C_0\;,\;C_1\;,\;C_2,\;for\;easiness.\;$$

1. The sum of Binomial Coefficients in the expansion of (1+x)n is 2n. $$\;Proof:\;we\;have,\;(1+x)^n\;=\;^nC_0\;+\;^nC_1x\;+\;^nC_2x^2\;+\;.\;.\;.\;.\;.\;+\;^nC_nx^n$$ $$Putting\;x=1,\;we\;get,\;$$ $$(1+1)^n\;=\;C_0\;+C_1+C_2+C_3+.\;.\;.\;.+C_n\;=\;2^n\;$$
2. The sum of the Coefficients of odd terms is equal to the sum of the coefficients of the even terms each being equal to 2n-1. $$\;i.e.\;C_0+C_2+C_4+\;.\;.\;.\;.\;.\;=\;C_1+C_3+C_5\;+\;.\;.\;.\;.\;=\;2^{n-1}$$ $$Putting\;x=1\;and\;x=-1\;we\;get,\;$$ $$\;C_0\;+C_1+C_2+C_3+.\;.\;.\;.+C_n\;=\;2^n\;.\;.\;.\;.\;.\;.(1)$$ $$\;and\;0\;=\;C_0-C_1+C_2\;.\;.\;.\;.\;.\;+\;(-1)^nC_n\;.\;.\;.\;.\;.(2)\;$$ Adding them (1) and (2) ,we get, $$2^n=2(C_0+C_2+C_4+\;.\;.\;.\;.\;.\;.)\;$$ $$\therefore\;C_0+C_2+C_4+.\;.\;.\;.\;=\;2^{(n-1)}$$ Similarly subtracting 2 from 1 we get, $$\;C_1+C_3+C_5+.\;.\;.\;.\;=\;2^{(n-1)}$$$$\therefore\;C_0+C_2+C_4+.\;.\;.\;.\;=\;C_1+C_3+C_5+\;.\;.\;.\;=\;2^{(n-1)}$$
3. In the expansion of (1+x)n the coefficients of terms equidistance from the beginning and the end are equal. Proof: The coefficient of (r+1)th term from the begining isnCr . Total number of expansion is n +1. $$\therefore\;(r+1)^{th}\;term\;from\;the\;end\;has\;(n+1)\;-\;(r+1)\;or\;n-r\;terms\;before\;it\;.$$ $$\;(r+1)^{th}\;term\;from\;the\;end\;=\;(n-r+1)^{th}\;term\;from\;the\;beginning\;$$ Its Coefficient is,nCn-r$$\;Since\;^nC_r\;=\;^nC_{n-r},\;the\;coefficients\;are\;equal\;.$$

#### Applications of Binomial Expansions

The Binomial theorem has different essential application. The most useful among them that you must know is the determination of approximate values of certain algebraic as well as arithmetical quantities and sums of some infinite series.

As an application of the binomial theorem,we can use the relation,

$$\;(1+x)^n\;=\;^nC_0\;+\;^nC_1x\;+\;^nC_2x^2\;+\;.\;.\;.\;.\;.\;+\;^nC_nx^n$$

$$\;i.e.\;(1+x)^n\;=\;1\;+\;nx\;+\;n(n-1)\frac{x^2}{2!}\;+\;\frac{n(n-1)(n-2)x^3}{3!}\;.\;.\;.\;.\;+\;\frac{n(n-1)(n-2)\;.\;.\;.(n-r-1)}{r!}.x^r\;+\;.\;.\;.\;.$$

which terminates after n+1 terms and when n is any positive integer. But when is any real number different from positive integer, the expansion does not terminate and it is valid, only if |x|<1 , and this expansion is known as binomial series. The proof is beyond the course. Also,

$$\;(1-x)^{-n}\;=\;1+nx+\;\frac{n(n+1)}{2!}.x^2+\frac{n(n+1)(n+2)}{3!}.x^3+\;.\;.\;.\;.\;\;\;\;\;\;\;\;\;\;\;(|x|<1)$$

Application of Binomial Theorem to find the nth root:

Let us suppose we have to find the nth root of a number N. First of all, we find a number 'á' such that an is as near to N as possible and write N=an+x. Here x will be positive if N>an and negative if N<an, but |x|/a will be a small fraction. Now we can apply the binomial theorem to expand N1/n (an+x)1/n i.e. a(1+x/an)1/nas shown in te example below.

Example:Find the cube root of 999, correct to 3 places of decimal.

Solution:The cube root of 999 = (999)3 = (1000-1)1/3

$$\;=\;(10^3-3)^{1/3}\;=\;(10^3)^{1/3}\bigg[\;1-\frac{1}{10^3}\bigg]^{1/3}$$

$$\;=\;10\bigg[1-\frac{1}{10^3}\bigg]\;=\;10(1-0.001)^{1/3}$$

$$\;10\;\Bigg[1+\frac{1}{3}(-0.001)\;+\;\frac{\frac{1}{3}\;\bigg(\frac{1}{3}-1\bigg)}{2!}.(-0.001)^2+\;.\;.\;.\;.\;.\;.\Bigg]$$

$$\;=10\bigg[1-0.000333-\frac{1}{9}(0.000001)+\;.\;.\;.\;.\;.\;.\bigg]$$

$$\;=10[1-0.000333-0.0000001+\;.\;.\;.\;.]$$

$$\;=\;10(0.9996669)\;approx.$$

$$\;=\;9.996669$$

$$\therefore\;\sqrt[3]{999}\;=\;9.997\;correct\;to\;3\;places\;decimal\;.\;$$

Application of Binomial Theorem to find the Summation of Infinite series:

Example: $$\;Find\;the\;value\;of\;1+\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5.7}{4.8.12}\;+\;.\;.\;.\;.\;.$$

Solution:

$$\;=1+\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5.7}{4.8.12}\;+\;.\;.\;.\;.\;.$$

$$\;=\;1+\frac{3}{2}.\frac{1}{2}+\{\frac{\frac{3}{2}.\frac{5}{2}}{2!}.\bigg(\frac{1}{2}\bigg)^2+\{\frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!}.\bigg(\frac{1}{2}\bigg)^3+\;.\;.\;.$$

$$\;=\;1+\frac{-3}{2}.\frac{-1}{2}+\{\frac{\frac{-3}{2}.\frac{-5}{2}}{2!}.\bigg(-\frac{1}{2}\bigg)^2+\{\frac{\frac{-3}{2}.\frac{-5}{2}.\frac{-7}{2}}{3!}.\bigg(-\frac{1}{2}\bigg)^3+\;.\;.\;.$$

$$\;Comparing\;with\;1+nx+\frac{n(n-1)}{2!}x^2\;+\;\;.\;.\;.\;.\;$$

$$\;=\;\bigg[\;1+\bigg(\frac{-1}{2}\bigg)\bigg]^{-\frac{3}{2}}$$

$$\;=\;\bigg[\;1-\frac{1}{2}\bigg]^{-\frac{3}{2}}\;=\bigg(\frac{1}{2}\bigg)^{-\frac{3}{2}}\;=\;2^{-\frac{3}{2}}\;=(2^3)^{\frac{1}{2}}\;=\sqrt{2^3}\;=\sqrt{8}\;$$

$$\therefore\;1+\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5.7}{4.8.12}\;+\;.\;.\;.\;.\;.\;=\;\sqrt{8}$$

Taken reference from

( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )

##### Things to remember
• The sum of Binomial Coefficients in the expansion of (1+x)n is 2n
•  The sum of the Coefficients of odd terms is equal to the sum of the coefficients of the even terms each being equal to 2n-1
• In the expansion of (1+x)n the coefficients of terms equidistance from the beginning and the end are equal.
• The Binomial theorem has different essential application. The most useful among them that you must know is the determination of approximate values of certain algebraic as well as arithmetical quantities and sums of some infinite series.

As an application of the binomial theorem,we can use the relation,

$$\;(1+x)^n\;=\;^nC_0\;+\;^nC_1x\;+\;^nC_2x^2\;+\;.\;.\;.\;.\;.\;+\;^nC_nx^n$$

$$\;i.e.\;(1+x)^n\;=\;1\;+\;nx\;+\;n(n-1)\frac{x^2}{2!}\;+\;\frac{n(n-1)(n-2)x^3}{3!}\;.\;.\;.\;.\;+\;\frac{n(n-1)(n-2)\;.\;.\;.(n-r-1)}{r!}.x^r\;+\;.\;.\;.\;.$$

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