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1

AIEEE 2005

MCQ (Single Correct Answer)
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.

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

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

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

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

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

$$ \Rightarrow$$$$ {}^{56}{C_4}$$
2

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$${S_n}$$

$$\therefore$$ $${{{t_n}} \over {{S_n}}} = {n \over 2}$$
3

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

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

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

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

= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$
4

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,

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

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

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

Questions Asked from Mathematical Induction and Binomial Theorem

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