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1

### AIEEE 2005

If the coefficient of $${x^7}$$ in $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ equals the coefficient of $${x^{ - 7}}$$ in $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$, then $$a$$ and $$b$$ satisfy the relation
A
$$a - b = 1$$
B
$$a + b = 1$$
C
$${a \over b} = 1$$
D
$$ab = 1$$

## Explanation

General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x7,

$$\Rightarrow$$ 22 - 3r = 7

$$\Rightarrow$$ r = 5

So coefficient of x7 = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is Tr+1.

Tr+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x-7,

11 - 3r = -7

$$\Rightarrow$$ r = 6

$$\therefore$$ Coefficient of x-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

According to question,

Coefficient of x7 = Coefficient of x-7

$$\Rightarrow$$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$\Rightarrow$$ $$ab = 1$$
2

### AIEEE 2005

If $$A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$ and $$I = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right],$$ then which one of the following holds for all $$n \ge 1,$$ by the principle of mathematical induction
A
$${A^n} = nA - \left( {n - 1} \right){\rm I}$$
B
$${A^n} = {2^{n - 1}}A - \left( {n - 1} \right){\rm I}$$
C
$${A^n} = nA + \left( {n - 1} \right){\rm I}$$
D
$${A^n} = {2^{n - 1}}A + \left( {n - 1} \right){\rm I}$$

## Explanation

Given $$A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$

$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$$

and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$$

So we can say $${A^n}$$ = $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$

Now $$nA - \left( {n - 1} \right){\rm I}$$

= $$\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$$ - $$\left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right]$$

= $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$ = $${A^n}$$

$$\therefore$$ $${A^n} = nA - \left( {n - 1} \right){\rm I}$$
3

### AIEEE 2005

The value of $$\,{}^{50}{C_4} + \sum\limits_{r = 1}^6 {^{56 - r}} {C_3}$$ is
A
$${}^{55}{C_4}$$
B
$${}^{55}{C_3}$$
C
$${}^{56}{C_3}$$
D
$${}^{56}{C_4}$$

## Explanation

Given, $${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}}$$

$$\Rightarrow$$ $${}^{50}{C_4}$$ + $${}^{55}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{50}{C_3}$$

Arrange those this way

$$\Rightarrow$$ $${}^{50}{C_4}$$ + $${}^{50}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

We know this formula [ $${{}^n{C_r}}$$ + $${{}^n{C_{r - 1}}}$$ = $${{}^{n + 1}{C_r}}$$ ] which is used to solve this problem.

$$\Rightarrow$$ $${}^{51}{C_4}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$\Rightarrow$$ $${}^{52}{C_4}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$\Rightarrow$$$${}^{53}{C_4}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$\Rightarrow$$$${}^{54}{C_4}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$\Rightarrow$$$${}^{55}{C_4}$$ + $${}^{55}{C_3}$$

$$\Rightarrow$$$${}^{56}{C_4}$$
4

### AIEEE 2004

If $${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}} \,\,and\,\,{t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}},\,}$$then $${{{t_{ n}}} \over {{S_n}}}$$ is equal to
A
$${{2n - 1} \over 2}$$
B
$${1 \over 2}n - 1$$
C
n - 1
D
$${1 \over 2}n$$

## Explanation

$${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$$

=$${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$$

$${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$$

= $${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}$$.........(1)

We can write $${t_n}$$ by rearranging like this,

$${t_n}$$ = $${n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}$$

= $${n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}$$.........(2)

[as $${{}^n{C_0}}$$ = $${{}^n{C_n}}$$, $${{}^n{C_1}}$$ = $${{}^n{C_{n - 1}}}$$......]

By adding (1) and (2) we get,

$$2{t_n}$$ = $${n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}$$

= $$n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]$$

= n$${S_n}$$

$$\therefore$$ $${{{t_n}} \over {{S_n}}} = {n \over 2}$$

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