First slide

Binomial theorem for positive integral Index

Question

If ${(1 + x + x^{2})}^{20} = a_{0} + a_{1} x + a_{2} x^{2} \dots + a_{40} x^{40},$ then answer the following questions.

Moderate

Question

The value of $a_{0} + a_{1} + a_{2} + \dots + a_{19}$ is

A

$\frac{1}{2} (9^{10} + a_{20})$
B

$\frac{1}{2} (9^{10} - a_{20})$
C

$\frac{9^{10}}{2}$
D

none of these

Solution

$Let, {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{r}$ ………(1)
Replacing x by 1/x, we get
$\begin{array}{l} {(1 + \frac{1}{x} + \frac{1}{x^{2}})}^{20} = \sum_{r = 0}^{40} a_{r} {(\frac{1}{x})}^{r} \\ or {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{40 - r} . . . . . . . . (2) \end{array}$
Since (1) and (2) are same series, coefficient of x^r in (1) : coefficient of x^r in (2)
$\Rightarrow a_{r} = a_{40 - r}$
In (1), putting x = 1, we get
$\begin{array}{l} 3^{20} = a_{0} + a_{1} + a_{2} + \dots + a_{40} \\ = (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} + (a_{21} + a_{n + 2} + \dots + a_{40}) \\ = 2 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} (∵ a_{r} = a_{40 - r}) \\ or a_{0} + a_{1} + a_{2} + \dots + a_{19} = \frac{1}{2} (3^{20} - a_{20}) = \frac{1}{2} (9^{10} - a_{20}) \end{array}$
Also,
$\begin{array}{l} a_{0} + 3 a_{1} + 5 a_{2} + \dots + 81 a_{40} \\ = (a_{0} + 81 a_{40}) + (3 a_{1} + 79 a_{39}) + \dots + (39 a_{19} + 43 a_{21}) + 41 a_{20} \end{array}$
$\begin{matrix} = 82 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + 41 a_{20} \\ = 41 (9^{10} - a_{20}) + 41 a_{20} \\ = 41 \times 3^{20} \end{matrix}$
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + \dots$ suggests that we have to multiply the two expansions.
Replacing r by - 1/x in (1), we get
$\begin{array}{l} {(1 - \frac{1}{x} + \frac{1}{x^{2}})}^{20} = a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} - \dots + \frac{a_{40}}{x^{40}} \\ \Rightarrow {(1 - x + x^{2})}^{20} = a_{0} x^{40} - a_{1} x^{39} + a_{2} x^{38} - \dots + a_{40} . . . . . . (3) \end{array}$
Clearly,
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} + \dots + a_{40}^{2}$ is the coefficient of x⁴⁰ in ${(1 + x + x^{2})}^{20} {(1 - x + x^{2})}^{20}$
$= Coefficient of x^{40} in {(1 + x^{2} + x^{4})}^{20}$
In ${(1 + x^{2} + x^{4})}^{20}$ , replace x² by y, then the coefficient of y²⁰ in ${(1 + y + y^{2})}^{20}$ is a₂₀.
$Hence, a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{40}^{2} = a_{20}$
$\begin{array}{l} or (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} + (- a_{21}^{2} + \dots + a_{40}^{2}) = a_{20} \\ or 2 (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} = a_{20} \\ or a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2} = \frac{a_{20}}{2} [1 - a_{20}] \end{array}$

Question

The value of $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}$ is

Solution

$Let, {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{r}$ ………(1)
Replacing x by 1/x, we get
$\begin{array}{l} {(1 + \frac{1}{x} + \frac{1}{x^{2}})}^{20} = \sum_{r = 0}^{40} a_{r} {(\frac{1}{x})}^{r} \\ or {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{40 - r} . . . . . . . . (2) \end{array}$
Since (1) and (2) are same series, coefficient of x^r in (1) : coefficient of x^r in (2)
$\Rightarrow a_{r} = a_{40 - r}$
In (1), putting x = 1, we get
$\begin{array}{l} 3^{20} = a_{0} + a_{1} + a_{2} + \dots + a_{40} \\ = (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} + (a_{21} + a_{n + 2} + \dots + a_{40}) \\ = 2 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} (∵ a_{r} = a_{40 - r}) \\ or a_{0} + a_{1} + a_{2} + \dots + a_{19} = \frac{1}{2} (3^{20} - a_{20}) = \frac{1}{2} (9^{10} - a_{20}) \end{array}$
Also,
$\begin{array}{l} a_{0} + 3 a_{1} + 5 a_{2} + \dots + 81 a_{40} \\ = (a_{0} + 81 a_{40}) + (3 a_{1} + 79 a_{39}) + \dots + (39 a_{19} + 43 a_{21}) + 41 a_{20} \end{array}$
$\begin{matrix} = 82 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + 41 a_{20} \\ = 41 (9^{10} - a_{20}) + 41 a_{20} \\ = 41 \times 3^{20} \end{matrix}$
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + \dots$ suggests that we have to multiply the two expansions.
Replacing r by - 1/x in (1), we get
$\begin{array}{l} {(1 - \frac{1}{x} + \frac{1}{x^{2}})}^{20} = a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} - \dots + \frac{a_{40}}{x^{40}} \\ \Rightarrow {(1 - x + x^{2})}^{20} = a_{0} x^{40} - a_{1} x^{39} + a_{2} x^{38} - \dots + a_{40} . . . . . . (3) \end{array}$
Clearly,
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} + \dots + a_{40}^{2}$ is the coefficient of x⁴⁰ in ${(1 + x + x^{2})}^{20} {(1 - x + x^{2})}^{20}$
$= Coefficient of x^{40} in {(1 + x^{2} + x^{4})}^{20}$
In ${(1 + x^{2} + x^{4})}^{20}$ , replace x² by y, then the coefficient of y²⁰ in ${(1 + y + y^{2})}^{20}$ is a₂₀.
$Hence, a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{40}^{2} = a_{20}$
$\begin{array}{l} or (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} + (- a_{21}^{2} + \dots + a_{40}^{2}) = a_{20} \\ or 2 (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} = a_{20} \\ or a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2} = \frac{a_{20}}{2} [1 - a_{20}] \end{array}$

Question

The value of $a_{0} + 3 a_{1} + 5 a_{2} + \dots + 81 a_{40}$ is

Solution

$Let, {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{r}$ ………(1)
Replacing x by 1/x, we get
$\begin{array}{l} {(1 + \frac{1}{x} + \frac{1}{x^{2}})}^{20} = \sum_{r = 0}^{40} a_{r} {(\frac{1}{x})}^{r} \\ or {(1 + x + x^{2})}^{20} = \sum_{r = 0}^{40} a_{r} x^{40 - r} . . . . . . . . (2) \end{array}$
Since (1) and (2) are same series, coefficient of x^r in (1) : coefficient of x^r in (2)
$\Rightarrow a_{r} = a_{40 - r}$
In (1), putting x = 1, we get
$\begin{array}{l} 3^{20} = a_{0} + a_{1} + a_{2} + \dots + a_{40} \\ = (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} + (a_{21} + a_{n + 2} + \dots + a_{40}) \\ = 2 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + a_{20} (∵ a_{r} = a_{40 - r}) \\ or a_{0} + a_{1} + a_{2} + \dots + a_{19} = \frac{1}{2} (3^{20} - a_{20}) = \frac{1}{2} (9^{10} - a_{20}) \end{array}$
Also,
$\begin{array}{l} a_{0} + 3 a_{1} + 5 a_{2} + \dots + 81 a_{40} \\ = (a_{0} + 81 a_{40}) + (3 a_{1} + 79 a_{39}) + \dots + (39 a_{19} + 43 a_{21}) + 41 a_{20} \end{array}$
$\begin{matrix} = 82 (a_{0} + a_{1} + a_{2} + \dots + a_{19}) + 41 a_{20} \\ = 41 (9^{10} - a_{20}) + 41 a_{20} \\ = 41 \times 3^{20} \end{matrix}$
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - a_{3}^{2} + \dots$ suggests that we have to multiply the two expansions.
Replacing r by - 1/x in (1), we get
$\begin{array}{l} {(1 - \frac{1}{x} + \frac{1}{x^{2}})}^{20} = a_{0} - \frac{a_{1}}{x} + \frac{a_{2}}{x^{2}} - \dots + \frac{a_{40}}{x^{40}} \\ \Rightarrow {(1 - x + x^{2})}^{20} = a_{0} x^{40} - a_{1} x^{39} + a_{2} x^{38} - \dots + a_{40} . . . . . . (3) \end{array}$
Clearly,
$a_{0}^{2} - a_{1}^{2} + a_{2}^{2} + \dots + a_{40}^{2}$ is the coefficient of x⁴⁰ in ${(1 + x + x^{2})}^{20} {(1 - x + x^{2})}^{20}$
$= Coefficient of x^{40} in {(1 + x^{2} + x^{4})}^{20}$
In ${(1 + x^{2} + x^{4})}^{20}$ , replace x² by y, then the coefficient of y²⁰ in ${(1 + y + y^{2})}^{20}$ is a₂₀.
$Hence, a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots + a_{40}^{2} = a_{20}$
$\begin{array}{l} or (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} + (- a_{21}^{2} + \dots + a_{40}^{2}) = a_{20} \\ or 2 (a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2}) + a_{20}^{2} = a_{20} \\ or a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - \dots - a_{19}^{2} = \frac{a_{20}}{2} [1 - a_{20}] \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App