First slide

Binomial theorem for positive integral Index

Question

An equation $a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = 0$ has roots $^{99} C_{0},^{99} C_{1},^{99} C_{2}, \dots,^{99} C_{99}$ .

Moderate

Question

The value of a₉₉ is equal to

A

2⁹⁸
B

2⁹⁹
C

-2⁹⁹
D

none of these

Solution

$a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = 0 {has roots}^{99} C_{0},^{99} C_{1},^{99} C_{2}, \dots,^{99} C_{99}$
$or a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = (x -^{99} C_{0}) (x -^{99} C_{1}) (x -^{99} C_{2}) \dots (x -^{99} C_{99})$
Now, sum of roots is
$^{99} C_{0} +^{99} C_{1} +^{99} C_{2} + \dots +^{99} C_{99} = - \frac{a_{99}}{coefficient of x^{100}}$
$or a_{99} = - 2^{99}$
Also, sum of product of roots taken two at a time is
$\frac{a_{99}}{coefficient of x^{100}}$
$\begin{array}{l} ∴ \underset{0 \leq i < j \leq 99}{\sum \sum}_{i = 0}^{99} {C_{i}}^{99} C_{j} = = \frac{(\sum_{i = 0}^{99} \sum_{j = 0}^{99}^{99} C_{i}^{99} C_{j}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \\ = \frac{2^{99} 2^{99} - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{(\sum_{i = 0}^{99} C_{i} 2^{99}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{2^{198} -^{198} C_{99}}{2} \end{array}$
$\begin{matrix} {(^{99} C_{0})}^{2} + {(^{99} C_{1})}^{2} + \dots + {(^{99} C_{99})}^{2} \\ = {(^{99} C_{0} +^{99} C_{1} +^{99} C_{2} \dots +^{99} C_{99})}^{2} - 2 \\ = {(- a_{99})}^{2} - 2 a_{98} \underset{0 \leq i < j \leq 99}{\sum \sum}^{99} {C_{i}}^{99} C_{j} \\ = a_{99}^{2} - 2 a_{98} \end{matrix}$

Question

The value of a₉₈is

A

$\frac{2^{198} -^{198} C_{99}}{2}$
B

$\frac{2^{198} +^{198} C_{99}}{2}$
C

$2^{99} -^{99} C_{49}$
D

none of these

Solution

$a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = 0 {has roots}^{99} C_{0},^{99} C_{1},^{99} C_{2}, \dots,^{99} C_{99}$
$or a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = (x -^{99} C_{0}) (x -^{99} C_{1}) (x -^{99} C_{2}) \dots (x -^{99} C_{99})$
Now, sum of roots is
$^{99} C_{0} +^{99} C_{1} +^{99} C_{2} + \dots +^{99} C_{99} = - \frac{a_{99}}{coefficient of x^{100}}$
$or a_{99} = - 2^{99}$
Also, sum of product of roots taken two at a time is
$\frac{a_{99}}{coefficient of x^{100}}$
$\begin{array}{l} ∴ \underset{0 \leq i < j \leq 99}{\sum \sum}_{i = 0}^{99} {C_{i}}^{99} C_{j} = = \frac{(\sum_{i = 0}^{99} \sum_{j = 0}^{99}^{99} C_{i}^{99} C_{j}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \\ = \frac{2^{99} 2^{99} - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{(\sum_{i = 0}^{99} C_{i} 2^{99}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{2^{198} -^{198} C_{99}}{2} \end{array}$
$\begin{matrix} {(^{99} C_{0})}^{2} + {(^{99} C_{1})}^{2} + \dots + {(^{99} C_{99})}^{2} \\ = {(^{99} C_{0} +^{99} C_{1} +^{99} C_{2} \dots +^{99} C_{99})}^{2} - 2 \\ = {(- a_{99})}^{2} - 2 a_{98} \underset{0 \leq i < j \leq 99}{\sum \sum}^{99} {C_{i}}^{99} C_{j} \\ = a_{99}^{2} - 2 a_{98} \end{matrix}$

Question

The value of ${(^{99} C_{0})}^{2} + {(^{99} C_{1})}^{2} + \dots + {(^{99} C_{99})}^{2}$ is equal to

A

$2 a_{98} - a_{99}^{2}$
B

$a_{99}^{2} - a_{98}$
C

$a_{99}^{2} - 2 a_{98}$
D

none of these

Solution

$a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = 0 {has roots}^{99} C_{0},^{99} C_{1},^{99} C_{2}, \dots,^{99} C_{99}$
$or a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{99} x^{99} + x^{100} = (x -^{99} C_{0}) (x -^{99} C_{1}) (x -^{99} C_{2}) \dots (x -^{99} C_{99})$
Now, sum of roots is
$^{99} C_{0} +^{99} C_{1} +^{99} C_{2} + \dots +^{99} C_{99} = - \frac{a_{99}}{coefficient of x^{100}}$
$or a_{99} = - 2^{99}$
Also, sum of product of roots taken two at a time is
$\frac{a_{99}}{coefficient of x^{100}}$
$\begin{array}{l} ∴ \underset{0 \leq i < j \leq 99}{\sum \sum}_{i = 0}^{99} {C_{i}}^{99} C_{j} = = \frac{(\sum_{i = 0}^{99} \sum_{j = 0}^{99}^{99} C_{i}^{99} C_{j}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \\ = \frac{2^{99} 2^{99} - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{(\sum_{i = 0}^{99} C_{i} 2^{99}) - \sum_{i = 0}^{99} {(^{99} C_{i})}^{2}}{2} \\ = \frac{2^{198} -^{198} C_{99}}{2} \end{array}$
$\begin{matrix} {(^{99} C_{0})}^{2} + {(^{99} C_{1})}^{2} + \dots + {(^{99} C_{99})}^{2} \\ = {(^{99} C_{0} +^{99} C_{1} +^{99} C_{2} \dots +^{99} C_{99})}^{2} - 2 \\ = {(- a_{99})}^{2} - 2 a_{98} \underset{0 \leq i < j \leq 99}{\sum \sum}^{99} {C_{i}}^{99} C_{j} \\ = a_{99}^{2} - 2 a_{98} \end{matrix}$

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