Notes on Examples on Binomial Theorem (Binomial Coefficients, Application of Binomial Theorem,General Term) | Grade 12 > Mathematics > Binomial Theorem | KULLABS.COM

Examples on Binomial Theorem (Binomial Coefficients, Application of Binomial Theorem,General Term)

Notes, Exercises, Videos, Tests and Things to Remember on Examples on Binomial Theorem (Binomial Coefficients, Application of Binomial Theorem,General Term)

Please scroll down to get to the study materials.

Registration open for Special Scholarships

  • Note
  • Things to remember

Example 1:

Expand (2a+b)­ by the binomial theorem.

Solution:

With 2a as the first term and b as the second term in the binomial expression to be expanded, we get

(2a+b)5

$$\;=(2a)^5+5(2a)^4.b+\frac{5.4}{2!}(2a)^3.b^2+\frac{5.4.3}{3!}(2a)^2.b^3+5.2a.b^4+b^5$$

$$\;=\;32a^5+80a^4.b+80a^3b^2+40a^2b^3+10ab^4+b^5$$

Example 2:

$$Find\;the\;seventh\;term\;in\;the\;expansion\;of\;(2x+\frac{1}{y})^{10}$$

Solution:

$$7^{th}\;term\;=t_7=t_{6+1}=C(10,6).(2x)^{10-6}.\bigg(\;\frac{1}{y}\bigg)^6$$

Example 3:

$$Find\;the\;general\;term\;in\;the\;expansion\;of\;\bigg(\;x^2+\frac{a^2}{x}\bigg)^5.\;Then\;find\;the\;coefficient\;of\;x.$$

Solution:

$$\;Let\;t_{r+1}\;be\;the\;general\;term\;in\;the\;expansion\;of\;\bigg(\;x^2+\frac{a^2}{x}\bigg)^5.$$

Then,

$$t_{r+1}=C(5,r)(x^2)^{5-r}.\frac{a^2}{x}^r$$

$$\;=C(5,r)\;x^{10-2r}.\frac{a^{2r}}{x^r}$$

$$\;=\;C(5,r)x^{10-3r}.a^{2r}$$

To get the coefficient of x,

$$10-3r=1$$

$$or,\;\;\;\;9=3r\;\;$$

$$\therefore\;\;\;\;r=3$$

$$\therefore\;t_4=4^{th}\;term\;contains\;x$$

$$\therefore\;Coefficient\;of\;x\;=\;C(5,3)a^6=10a^6$$

Example 4:

Find the middle term in the expansion of (2a+3x)­30.

Solution:

Here, the only number of terms in the expansion of (2a+3x)30 is 30+1 i.e. 31. So, there is only one mid term.

$$\therefore\;=midterm=\;t_{(30/2)+1}=t_{15+1}$$

$$=C(30,15)\;(2a)^{30-15}.(3x)^{15}.$$

$$=C(30,15)\;(2a)^{15}.(3x)^{15}.$$

$$=\frac{30!}{(15!)^2}\;(2a)^{15}.(3x)^{15}.$$

Hence the mid term of (2a+3x)30 is found to be

$$=\frac{30!}{(15!)^2}\;(2a)^{15}.(3x)^{15}.$$

Example 5:

$$Find\;the\;middle\;term\;in\;the\;expansion\;\bigg(1+\frac{x}{2}\bigg)^{15}.$$

Solution:

Here, the numbers of terms in the expansion is 15+1=16 which is even. So, there are two middle terms, The middle terms are t­(15+1)/2 and t­(15+1)/2+1 i.e. t­8 ­and t­9­.

$$t_8=t_{7+1}=C(15,7)\;(1)^{15-7}.\bigg(\frac{x}{2}\bigg)^{7}.$$

$$=\frac{15!}{8!.7!}\frac{x^7}{2^7}$$

$$and\;t_9=t_{8+1}=C(15,8)\;(1)^{15-8}.\bigg(\frac{x}{2}\bigg)^{8}.$$

$$=\frac{15!}{8!.7!}\frac{x^8}{2^8}$$

Example 6:

$$Find\;the\;term\;independent\;of\;x\;in\;the\;expansion\;of\;\bigg(x+\frac{1}{3x^2}\bigg)^12$$

Solution:

Let t­r+1 be the term independent of x.

$$Then,t_{r+1}=C(12,r)\;(x)^{12-r}.\bigg(\frac{1}{3x^2}\bigg)^{r}.$$

$$\;=\;(-1)^r\frac{C(12,r)}{3^r}.x^{12-r}\frac{1}{x^{2r}}$$

$$\;=\; \frac{(-1)^r.C(12,r)}{3^r}.x^{12-3r}$$

For the term idependent of x,

$$12-3=0$$

$$\therefore\;r=4$$

$$\therefore\;t_5\;is\;the\;term\;independent\;of\;x$$

$$and\;t_5=\;\frac{(-1)^5.C(12,5)}{3^5}=\;frac{12\times\;11\times\;10\times\;9}{4\times\;3\times\;2\;times\;1\times\;81}=\frac{55}{9}$$

Example 7:

In the expansion of (1+x)­43 , coefficient of (2r+1)th term and (r+2)th term are equal. Find r.

Solution:

$$t_{2r+1}=C(43,2r)\;1.x^{2r}.$$

$$\therefore\;coefficient\;of\;t_{2r+1}= C(43,2r)$$

$$Again\;for,\;t_{r+2}=C(43,r+1)\;1.x^{r+1}.$$

$$\therefore\;coefficient\;of\;;t_{r+2}=C(43,r+1)$$

According to the question,

$$C(43,2r)=C(43,r+1)$$

$$\rightarrow\;2r=r+1$$

$$\rightarrow\;r=1$$

$$Also\;2r+r+1=43$$

$$\rightarrow\;3r=42$$

$$\therefore\;r=14$$

$$\therefore\;r=1\;or\;14$$

Example 8:

$$If\;(1+x)^n=C_0+C_1x+C_2x^2+…\;…\;…+C_nx^n\;Prove\;that\;$$

$$(a)\;C_1+2C_2+3C_3+…\;…\;..+nC_n=n2^{n-1}$$

$$(b)\;C_0C_2+C_1C_3+C_2C_4+\;…\;+C_{n-2}C_n=\frac{2n!}{(n-2)(n+2)!}$$

Solution:

$$(a)\;C_1+2C_2+3C_3+…\;…\;..+nC_n=n2^{n-1}$$

$$L.H.S= C_1+2C_2+3C_3+…\;…\;..+nC_n$$

$$\;\;=n+2n\frac{n-1}{2!}+\frac{3n(n-1)(n-2)}{3!}+\;…\;+\;n.1$$

$$\;\;=n\bigg(1+\frac{n-1}{1}+\frac{(n-1)(n-2)}{2!}+\;…\;+\;1\bigg)$$

$$\;\;=n(1+1)^n=n2^{2n-1}$$

$$(b)\;C_0C_2+C_1C_3+C_2C_4+\;…\;+C_{n-2}C_n=\frac{2n!}{(n-2)(n+2)!}$$

We have,

$$(1+x)^n=C_0+C_1x+C_2x^2+…\;…\;…+C_nx^n\;$$

$$And,\;\; (x+1)^n=C_0x^n+C_1x^{n-1}+C_2x^{n-2}+…\;…\;…+C_n\;$$

Multiplying the above equations,

$$(1+x)^{2n}=( C_0+C_1x+C_2x^2+…\;…\;…+C_nx^n)( C_0x^n+C_1x^{n-1}+C_2x^{n-2}+…\;…\;…+C_n\;)$$

Since it is an identity, the coefficient of any power of x of the LHS should be equal to the coefficient of the same power of x of the RHS.

The coefficient of any power of xn-2 in the L. H. S. ,

$$\;=C(2n,n+2)=\frac{2n!}{(n-2)!(n+2)!}$$

The coefficient of any power of xn-2 in the R. H. S. ,

$$=C_0C_2+C_1C_3+C_2C_4+\;…\;+C_{n-2}C_n$$

$$\therefore\;C_0C_2+C_1C_3+C_2C_4+\;…\;+C_{n-2}C_n=\frac{2n!}{(n-2)(n+2)!}$$

Example 9:

Find the value of (1.03)-5 correct to three significant figures.

Solution:

We have,

$$(1.03)^{-5}=(1+0.03)^{-5}\;\;(\vert\;x\vert\;=0.03<1)$$

$$\;=1+(-5)(0.03)+\frac{(-5)(-6)}{2!}(0.03)^2+\frac{(-5)(-6)(-7)}{3!}(0.03)^3+\;…\;…$$

$$\;=1-0.15+0.0135-0.000945+\;…\;..\;…\;…$$

$$\;=0.862555=0.863.$$

Example 10:

$$Show\;that\;\sqrt{8}=1+\frac{3}{4}+\frac{3.5}{4.8}+\frac{3.5.7}{4.8.12}+\;…\;…\;…\;…$$

Solution:

$$Here,\;\sqrt{8}=\sqrt{2^3}$$

$$=2^{\frac{3}{2}}=\bigg(\frac{1}{2}\bigg)^{-\frac{3}{2}}=\bigg(1-\frac{1}{2}\bigg)^{-\frac{3}{2}}$$

Applying binomial theorem for any index, we have

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

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

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

Taken reference from

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



.

Questions and Answers

Click on the questions below to reveal the answers

0%

ASK ANY QUESTION ON Examples on Binomial Theorem (Binomial Coefficients, Application of Binomial Theorem,General Term)

No discussion on this note yet. Be first to comment on this note