Binomial Theorem
5.0 Important Results
5.0 Important Results
1. Suppose it is given that $${(a + x)^{10}}{ = ^{10}}{C_0}{a^{10}}{ + ^{10}}{C_1}{a^9}x + ...{ + ^{10}}{C_{10}}{x^{10}}\quad ...(1)$$
and we have to find the ${4^{th}}$ term from last, then we have to interchange the place of $a$ and $x$ in the given expression and find the ${4^{th}}$ term from beginning of the new expression using general term formulae i.e., $${(x + a)^{10}}{ = ^{10}}{C_0}{x^{10}}{ + ^{10}}{C_1}{x^9}a + ...{ + ^{10}}{C_{10}}{a^{10}}\quad ...(2)$$
The ${4^{th}}$ term of $(2)$ from beginning is ${}^{10}{C_3}{x^7}{a^3}$ which is also the ${4^{th}}$ term of $(1)$ from last.
2. If ${(\sqrt A + B)^n} = I + f$, where $I$ and $n$ are positive integers , $n$ being odd and $0 < f < 1$, then $$(I + f)f = {k^n}$$ where $A - {B^2} = k > 0$ and $\sqrt A - B < 1$.
If $n$ is an even integer, then $$(I + f)(1 - f) = {k^n}$$
3. $$\begin{equation} \begin{aligned} {(a + x)^n}{ = ^n}{C_0}{a^{n - 0}}{x^0}{ + ^n}{C_1}{a^{n - 1}}{x^1}{ + ^n}{C_2}{a^{n - 2}}{x^2} + ...{ + ^n}{C_r}{a^{n - r}}{x^r} + ...{ + ^n}{C_{n - 1}}a{x^{n - 1}}{ + ^n}{C_n}{a^0}{x^n}\quad ...(1) \\ {(a - x)^n}{ = ^n}{C_0}{a^{n - 0}}{( - x)^0}{ - ^n}{C_1}{a^{n - 1}}{x^1}{ + ^n}{C_2}{a^{n - 2}}{x^2} - ...{ + ^n}{C_r}{a^{n - r}}{( - x)^r} + ...{ + ^n}{C_{n - 1}}a{( - x)^{n - 1}}{ + ^n}{C_n}{a^0}{( - x)^n}\quad ...(2) \\\end{aligned} \end{equation} $$
Adding $(1)$ and $(2)$, we get
$$\begin{equation} \begin{aligned} {(a + x)^n} + {(a - x)^n} = 2{[^n}{C_0}{a^{n - 0}}{x^0}{ + ^n}{C_2}{a^{n - 2}}{x^2} + ...{ + ^n}{C_n}{a^0}{x^n} \\ {\text{ = 2}}\left[ {{\text{sum of terms at odd places}}} \right] \\\end{aligned} \end{equation} $$
Subtracting $(2)$ from $(1)$, we get $$\begin{equation} \begin{aligned} {(a + x)^n} - {(a - x)^n} = 2{[^n}{C_1}{a^{n - 1}}{x^1}{ + ^n}{C_3}{a^{n - 3}}{x^3} + ...{ + ^n}{C_{n - 1}}a{x^{n - 1}} \\ {\text{ = 2[Sum of terms at even places]}} \\\end{aligned} \end{equation} $$
| If $n$ is even | If $n$ is odd | |
| Number of terms in ${(a + x)^n} + {(a - x)^n}$ | $\frac{n}{2} + 1$ | $\frac{{n + 1}}{2}$ |
| Number of terms in ${(a + x)^n} - {(a - x)^n}$ | $\frac{n}{2}$ | $\frac{{n + 1}}{2}$ |
Question 5. Let $A$ be the sum of odd terms and $B$ be the sum of even terms in ${(x + a)^n}$. Prove that
(i) ${A^2} - {B^2} = {({x^2} - {a^2})^n}$
(ii) $4AB = {(x + a)^{2n}} - {(x - a)^{2n}}$
Solution: (i) $A$ = sum of odd terms and $B$ = sum of even terms
Therefore, $A+B$ = sum of odd terms + sum of even terms = ${(x + a)^n}$ ...$(1)$
Similarly, $A-B$ = ${(x - a)^n}$ ...$(2)$
Multiply $(1)$ and $(2)$, we get
$$\begin{equation} \begin{aligned} (A + B)(A - B) = {(x + a)^n}{(x - a)^n} \\ {A^2} - {B^2} = {({x^2} - {a^2})^n} \\\end{aligned} \end{equation} $$
(ii) From equation $(1)$, $$A + B = {(x + a)^n}$$
Squaring both sides, we get $${A^2} + {B^2} + 2AB = {(x + a)^{2n}}\quad ...(3)$$
From equation $(2)$, $$A - B = {(x - a)^n}$$
Squaring both sides, we get $${A^2} + {B^2} - 2AB = {(x - a)^{2n}}\quad ...(4)$$
Subtracting $(4)$ from $(3)$, we get $$\begin{equation} \begin{aligned} {A^2} + {B^2} + 2AB - {A^2} - {B^2} + 2AB = {(x - a)^{2n}} - {(x - a)^{2n}} \\ 4AB = {(x - a)^{2n}} - {(x - a)^{2n}} \\\end{aligned} \end{equation} $$
Question 6. Using binomial theorem prove that ${2^{3n}} - 7n - 1$ is divisible by $49$.
Solution: ${2^{3n}} - 7n - 1 \equiv {8^n} - 7n - 1$
Now, $$\begin{equation} \begin{aligned} {8^n} = {(1 + 7)^n} \\ {8^n} = {}^n{C_0} + {}^n{C_1}(7) + {}^n{C_2}{(7)^2} + ... \\ {8^n}{\text{ = }}{}^n{C_0} + {}^n{C_1}(7) + 49[I] \\ {8^n} - 7n - 1 = 49[I] \\\end{aligned} \end{equation} $$
So, ${2^{3n}} - 7n - 1$ is divisible by $49$.
Question 7. If $n$ is positive integer, then prove that the integral part of ${(7 + 4\sqrt 3 )^n}$ is an odd number.
Solution: Let $${(7 + 4\sqrt 3 )^n} = I + f\quad ...(1)$$ where $I$ and $f$ are its integral and fractional parts respectively. It means $0<f<1$. Now, $$\begin{equation} \begin{aligned} 0 < 7 - 4\sqrt 3 < 1 \\ 0 < {(7 - 4\sqrt 3 )^n} < 1 \\\end{aligned} \end{equation} $$
Let $\begin{equation} \begin{aligned} {(7 - 4\sqrt 3 )^n} = f'\quad ...(2) \\ \Rightarrow 0 < f' < 1 \\\end{aligned} \end{equation} $
Adding $(1)$ and $(2)$, we get $$\begin{equation} \begin{aligned} I + f + f' = {(7 + 4\sqrt 3 )^n} + {(7 - 4\sqrt 3 )^n} \\ {\text{ = }}2[{}^n{C_0}{7^n} + {}^n{C_2}{7^{n - 2}}{(4\sqrt 3 )^2} + ...] \\\end{aligned} \end{equation} $$
$I+f+f'=$ even integer means $f+f'$ must be an integer.
$0<f+f'<2$ implies that $f+f'=1$
Since, $I+1=$even integer
Therefore, $I$ is an odd integer.
