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

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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 )

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