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1

### AIEEE 2004

MCQ (Single Correct Answer)
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$${S_n}$$

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

### AIEEE 2004

MCQ (Single Correct Answer)
The coefficient of $${x^n}$$ in expansion of $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$ is
A
$${\left( { - 1} \right)^{n - 1}}n$$
B
$${\left( { - 1} \right)^n}\left( {1 - n} \right)$$
C
$${\left( { - 1} \right)^{n - 1}}{\left( {n - 1} \right)^2}$$
D
$$\left( {n - 1} \right)$$

## Explanation

Given $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$

= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$

$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$$

So coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}$$

General term of $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$$

$$\therefore$$ Term containing $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$$

So coefficient of $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

$$\therefore$$ coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

= $${}^n{C_n}.{\left( { - 1} \right)^n}$$ + $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$$

= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$
3

### AIEEE 2004

MCQ (Single Correct Answer)
The coefficient of the middle term in the binomial expansion in powers of $$x$$ of $${\left( {1 + \alpha x} \right)^4}$$ and $${\left( {1 - \alpha x} \right)^6}$$ is the same if $$\alpha$$ equals
A
$${3 \over 5}$$
B
$${10 \over 3}$$
C
$${{ - 3} \over {10}}$$
D
$${{ - 5} \over {3}}$$

## Explanation

For $${\left( {1 + \alpha x} \right)^4}$$ the middle term $${T_{{4 \over 2} + 1}}$$ = $${}^4{C_2}.{\alpha ^2}{x^2}$$

$$\therefore$$ Coefficient of middle term = $${}^4{C_2}.{\alpha ^2}$$

For $${\left( {1 - \alpha x} \right)^6}$$ the middle term $${T_{{6 \over 2} + 1}}$$ = $${}^6{C_3}.-{\alpha ^3}{x^3}$$

$$\therefore$$ Coefficient of middle term = $${}^6{C_3}.{-\alpha ^3}$$

$$\therefore$$ According to question,

$${}^4{C_2}.{\alpha ^2}$$ = $${}^6{C_3}.{-\alpha ^3}$$

$$\Rightarrow 6 = 20 \times - \alpha$$

$$\Rightarrow \alpha = - {3 \over {10}}$$
4

### AIEEE 2004

MCQ (Single Correct Answer)
Let $$S(K)$$ $$= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}.$$ Then which of the following is true
A
Principle of mathematical induction can be used to prove the formula
B
$$S\left( K \right) \Rightarrow S\left( {K + 1} \right)$$
C
$$S\left( K \right) \ne S\left( {K + 1} \right)$$
D
$$S\left( 1 \right)$$ is correct

## Explanation

Given $$S(K)$$ $$= 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $$\ne$$ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)2

Now let S(k) is true

$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k2

$$\Rightarrow$$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k2 + 2k +1

= 3 + (k + 1)2

$$\Rightarrow$$ S(k + 1) is true.

$$\therefore$$ S(k) $$\Rightarrow$$ S(k + 1)

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